Real numbers definition problem












2












$begingroup$


In the definition of the limit, real positive numbers are used as distance, so we can not use the limit in the definition of real numbers. the definition of the limit.
But dumb numbers are defined as the limit of the sequence of rational numbers. Is not this a logical problem?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 20:31








  • 1




    $begingroup$
    In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
    $endgroup$
    – fonini
    Jan 6 at 20:34


















2












$begingroup$


In the definition of the limit, real positive numbers are used as distance, so we can not use the limit in the definition of real numbers. the definition of the limit.
But dumb numbers are defined as the limit of the sequence of rational numbers. Is not this a logical problem?










share|cite|improve this question









$endgroup$












  • $begingroup$
    Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 20:31








  • 1




    $begingroup$
    In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
    $endgroup$
    – fonini
    Jan 6 at 20:34
















2












2








2


1



$begingroup$


In the definition of the limit, real positive numbers are used as distance, so we can not use the limit in the definition of real numbers. the definition of the limit.
But dumb numbers are defined as the limit of the sequence of rational numbers. Is not this a logical problem?










share|cite|improve this question









$endgroup$




In the definition of the limit, real positive numbers are used as distance, so we can not use the limit in the definition of real numbers. the definition of the limit.
But dumb numbers are defined as the limit of the sequence of rational numbers. Is not this a logical problem?







definition real-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 20:25









alwaystudentalwaystudent

263




263












  • $begingroup$
    Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 20:31








  • 1




    $begingroup$
    In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
    $endgroup$
    – fonini
    Jan 6 at 20:34




















  • $begingroup$
    Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
    $endgroup$
    – Dietrich Burde
    Jan 6 at 20:31








  • 1




    $begingroup$
    In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
    $endgroup$
    – fonini
    Jan 6 at 20:34


















$begingroup$
Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
$endgroup$
– Dietrich Burde
Jan 6 at 20:31






$begingroup$
Check out the "true" definitions at this question. For example using Dedekind cuts, not limits.
$endgroup$
– Dietrich Burde
Jan 6 at 20:31






1




1




$begingroup$
In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
$endgroup$
– fonini
Jan 6 at 20:34






$begingroup$
In short, the notion of "limit" used in the construction of the reals from the rationals is what is called "Cauchy convergence", and makes no circular reference to real numbers, so that there is no logical problem.
$endgroup$
– fonini
Jan 6 at 20:34












3 Answers
3






active

oldest

votes


















3












$begingroup$

It is kind of a problem. Not in the sense that we don't know how to fix it, but because the fix is so relatively uninteresting that it doesn't really feel worth the effort to write it down.



You can develop much of the theory of metric spaces and limits by considering distances that are positive rationals only. This will be enough to allow you to define Cauchy sequences of rationals and make the push forward towards the real numbers.



Afterwards you'll then want to do everything over with real distances, for full generality. In particular you will then prove that it makes no difference whether the $varepsilon$s and $delta$s in the definitions range over the positive rationals or the positive reals.



Alternatively, start by defining metrics and limits even more abstractly by saying that the metric should take values in some ordered field with the Archimedean property. Since $mathbb Q$ is such a field, you can get started, and eventually it will turn out that $mathbb R$ is also such a field, so you can reuse what you did from the beginning.



The "aesthetic" downside of the latter route is that it will turn out that all ordered fields with the Archimedean property are actually (isomorphic to) subfields of $mathbb R$, so all that generality will not have bought you anything in the end. Thus people don't usually care to present the theory of metric spaces in that way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
    $endgroup$
    – Henning Makholm
    Jan 6 at 21:07










  • $begingroup$
    In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:18










  • $begingroup$
    dumb number means irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:34



















1












$begingroup$

Here's, in modern language, how Cantor defined the real numbers.



A sequence of rational numbers $(q_n)$ is said to be Cauchy if, for every positive rational $varepsilon$, there exists $k$ such that, for all $m,n>k$, $|q_m-q_n|<varepsilon$.



A sequence $(q_n)$ is a zero-sequence if, for every positive rational $varepsilon$, there exists $k$ such that, for all $n>k$, $|q_n|<varepsilon$. Every zero-sequence is Cauchy.



Fact 1. The set $mathscr{C}$ of Cauchy rational sequences forms a commutative ring under the operations $(q_n)+(r_n)=(q_n+r_n)$ and $(q_n)(r_n)=(q_nr_n)$.



The proof of this consists in showing that $mathscr{C}$ is a subring of the ring of all rational sequences with componentwise addition and multiplication.



Fact 2. The set $mathscr{Z}$ of zero-sequences is a maximal ideal of $mathscr{C}$.



The proof is direct verification.



Fact 3. The field $mathbb{Q}$ of rational numbers can be embedded in the quotient ring $mathbb{R}=mathscr{C}/mathscr{Z}$, which is a field as well.



The embedding consists in associating to a rational number the corresponding constant (Cauchy) sequence and verifying that in this way we get a ring homomorphism.



If $q=(q_n)$ is a Cauchy rational sequence, I'll denote with $hat{q}$ its image in the quotient ring $mathbb{R}$.



Fact 4. The field $mathbb{R}$ can be ordered.



Define $hat{q}<hat{r}$ if (and only if) there exists $k$ such that, for all $n>k$, $q_n<r_n$. It is readily seen that this is not dependent on the particular representative chosen for $hat{q}$ and $hat{r}$. Also the properties of ordered field are easy to verify.



Fact 5. (Cantor-Dedekind property) Every nonempty upper bounded subset of $mathbb{R}$ has a supremum.



This is the trickiest part. A closed interval is a subset of $mathbb{R}$ of the form $[hat{q},hat{r}]={hat{p}inmathbb{R}:hat{q}lehat{p}lehat{r}}$ (where $hat{p}<hat{r}$). A nested chain of closed intervals is given by two sequences $(hat{p}_n)$ and $(hat{r}_n)$ such that




  1. for all $n$, $hat{p}_n<hat{r}_n$;

  2. for all $m,n$, if $mle n$ then $hat{p}_mlehat{p}_n$ and $hat{r}_mgehat{r}_n$;

  3. for every positive rational $varepsilon$, there exists $k$ for which $hat{r}_k-hat{p}_k<varepsilon$.


It is not difficult (albeit tedious) to prove that, denoting by $pi_n$ the $n$-th term of a rational sequence representing $hat{p}_n$, and similarly defining $rho_n$ with the upper bounds, then





  1. $pi=(pi_n)$ and $rho=(rho_n)$ are Cauchy rational sequences;


  2. $hat{pi}=hat{rho}$;


  3. $hat{pi}=hat{rho}in[hat{p}_n,hat{r}_n]$, for every $n$.


In simpler words, given a nested chain of closed intervals, there exists a (necessarily unique) element of $mathbb{R}$ that belongs to every interval in the chain.



Deriving from this the Cantor-Dedekind property is standard.





As you see, the above construction of $mathbb{R}$ never uses the concept of limit, but it is of course modeled on it. Now it is essentially trivial to show that a Cauchy sequence in $mathbb{R}$ is convergent (with the standard definitions).



The construction of $mathbb{R}$ with Dedekind cuts allows for an easier derivation of the Cantor-Dedekind property. On the other hand, defining the field operations and verifying that we indeed get a field is much more boring. With the Cantor construction the field operations come almost for free from basic ring theory, but we pay a price when showing the Cantor-Dedekind property.



Both constructions give “the same” object, because it can be shown that two ordered fields having the Cantor-Dedekind property are isomorphic. The order relation on such a field is unique (essentially because every positive element has a square root). Such a field also contains a copy of the rational field $mathbb{Q}$ and is Archimedean.



Thus it is not relevant how we build the real numbers. Once we show that there exists a construction of an ordered field satisfying the Cantor-Dedekind property, we can free ourselves from the details of the construction and only use the properties of this field.



Now defining metric spaces with a metric taking its values in $mathbb{R}$ is not a problem any longer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:19












  • $begingroup$
    @alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
    $endgroup$
    – egreg
    Jan 7 at 15:21










  • $begingroup$
    dumb number used for irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:29










  • $begingroup$
    @alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
    $endgroup$
    – egreg
    Jan 7 at 15:47



















0












$begingroup$

One way of defining the real numbers is to say that they form an ordered field in which every non-empty subset which is bounded above has a least upper bound.



That then begs the question of whether such a field actually exists. The construction using Dedekind cuts, or alternatively using Cauchy sequences of the rationals, is sufficient to prove existence, and in fact then the least upper bound property does what you need.



There is also the question of whether such a field is unique, and indeed it is.



It is a common feature of mathematical definitions that a property is defined, and then is shown to be equivalent to some other useful properties and also to imply others. There is work to be done whichever equivalent definition is chosen, and sometimes we choose the definition which is easiest to work with in the context we have in mind.





So it depends a bit on whether you are using limits to define the real numbers, or to construct them, and then show that the numbers you construct have the properties you need.



If you define the real numbers via Cauchy Sequences of the Rationals, then there is a logical step to show that Cauchy Sequences of the Reals behave as expected (as Henning Makholm has indicated). Whichever way you do it some care has to be taken - it is not as easy as it might at first appear to show that real numbers constructed as Dedekind Sections obey the usual rules of arithmetic. However, the details have been worked through, and efficient proofs are known. But usually much of this detail is taken for granted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
    $endgroup$
    – Mark Bennet
    Jan 6 at 21:14











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064360%2freal-numbers-definition-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

It is kind of a problem. Not in the sense that we don't know how to fix it, but because the fix is so relatively uninteresting that it doesn't really feel worth the effort to write it down.



You can develop much of the theory of metric spaces and limits by considering distances that are positive rationals only. This will be enough to allow you to define Cauchy sequences of rationals and make the push forward towards the real numbers.



Afterwards you'll then want to do everything over with real distances, for full generality. In particular you will then prove that it makes no difference whether the $varepsilon$s and $delta$s in the definitions range over the positive rationals or the positive reals.



Alternatively, start by defining metrics and limits even more abstractly by saying that the metric should take values in some ordered field with the Archimedean property. Since $mathbb Q$ is such a field, you can get started, and eventually it will turn out that $mathbb R$ is also such a field, so you can reuse what you did from the beginning.



The "aesthetic" downside of the latter route is that it will turn out that all ordered fields with the Archimedean property are actually (isomorphic to) subfields of $mathbb R$, so all that generality will not have bought you anything in the end. Thus people don't usually care to present the theory of metric spaces in that way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
    $endgroup$
    – Henning Makholm
    Jan 6 at 21:07










  • $begingroup$
    In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:18










  • $begingroup$
    dumb number means irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:34
















3












$begingroup$

It is kind of a problem. Not in the sense that we don't know how to fix it, but because the fix is so relatively uninteresting that it doesn't really feel worth the effort to write it down.



You can develop much of the theory of metric spaces and limits by considering distances that are positive rationals only. This will be enough to allow you to define Cauchy sequences of rationals and make the push forward towards the real numbers.



Afterwards you'll then want to do everything over with real distances, for full generality. In particular you will then prove that it makes no difference whether the $varepsilon$s and $delta$s in the definitions range over the positive rationals or the positive reals.



Alternatively, start by defining metrics and limits even more abstractly by saying that the metric should take values in some ordered field with the Archimedean property. Since $mathbb Q$ is such a field, you can get started, and eventually it will turn out that $mathbb R$ is also such a field, so you can reuse what you did from the beginning.



The "aesthetic" downside of the latter route is that it will turn out that all ordered fields with the Archimedean property are actually (isomorphic to) subfields of $mathbb R$, so all that generality will not have bought you anything in the end. Thus people don't usually care to present the theory of metric spaces in that way.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
    $endgroup$
    – Henning Makholm
    Jan 6 at 21:07










  • $begingroup$
    In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:18










  • $begingroup$
    dumb number means irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:34














3












3








3





$begingroup$

It is kind of a problem. Not in the sense that we don't know how to fix it, but because the fix is so relatively uninteresting that it doesn't really feel worth the effort to write it down.



You can develop much of the theory of metric spaces and limits by considering distances that are positive rationals only. This will be enough to allow you to define Cauchy sequences of rationals and make the push forward towards the real numbers.



Afterwards you'll then want to do everything over with real distances, for full generality. In particular you will then prove that it makes no difference whether the $varepsilon$s and $delta$s in the definitions range over the positive rationals or the positive reals.



Alternatively, start by defining metrics and limits even more abstractly by saying that the metric should take values in some ordered field with the Archimedean property. Since $mathbb Q$ is such a field, you can get started, and eventually it will turn out that $mathbb R$ is also such a field, so you can reuse what you did from the beginning.



The "aesthetic" downside of the latter route is that it will turn out that all ordered fields with the Archimedean property are actually (isomorphic to) subfields of $mathbb R$, so all that generality will not have bought you anything in the end. Thus people don't usually care to present the theory of metric spaces in that way.






share|cite|improve this answer











$endgroup$



It is kind of a problem. Not in the sense that we don't know how to fix it, but because the fix is so relatively uninteresting that it doesn't really feel worth the effort to write it down.



You can develop much of the theory of metric spaces and limits by considering distances that are positive rationals only. This will be enough to allow you to define Cauchy sequences of rationals and make the push forward towards the real numbers.



Afterwards you'll then want to do everything over with real distances, for full generality. In particular you will then prove that it makes no difference whether the $varepsilon$s and $delta$s in the definitions range over the positive rationals or the positive reals.



Alternatively, start by defining metrics and limits even more abstractly by saying that the metric should take values in some ordered field with the Archimedean property. Since $mathbb Q$ is such a field, you can get started, and eventually it will turn out that $mathbb R$ is also such a field, so you can reuse what you did from the beginning.



The "aesthetic" downside of the latter route is that it will turn out that all ordered fields with the Archimedean property are actually (isomorphic to) subfields of $mathbb R$, so all that generality will not have bought you anything in the end. Thus people don't usually care to present the theory of metric spaces in that way.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 20:39

























answered Jan 6 at 20:32









Henning MakholmHenning Makholm

242k17308551




242k17308551












  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
    $endgroup$
    – Henning Makholm
    Jan 6 at 21:07










  • $begingroup$
    In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:18










  • $begingroup$
    dumb number means irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:34


















  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
    $endgroup$
    – Henning Makholm
    Jan 6 at 21:07










  • $begingroup$
    In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:18










  • $begingroup$
    dumb number means irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:34
















$begingroup$
I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
$endgroup$
– alwaystudent
Jan 6 at 21:05






$begingroup$
I already looked at it and it became clear that the rational numbers for the metric space are not sufficient, In fact, real numbers are the concept of geometry
$endgroup$
– alwaystudent
Jan 6 at 21:05






1




1




$begingroup$
@alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
$endgroup$
– Henning Makholm
Jan 6 at 21:07




$begingroup$
@alwaystudent: What do you think they are not sufficient for, and is this this on the critical path to defining the real numbers?
$endgroup$
– Henning Makholm
Jan 6 at 21:07












$begingroup$
In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
$endgroup$
– alwaystudent
Jan 7 at 15:18




$begingroup$
In fact, I do not remember my original proof, but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
$endgroup$
– alwaystudent
Jan 7 at 15:18












$begingroup$
dumb number means irrational number
$endgroup$
– alwaystudent
Jan 7 at 15:34




$begingroup$
dumb number means irrational number
$endgroup$
– alwaystudent
Jan 7 at 15:34











1












$begingroup$

Here's, in modern language, how Cantor defined the real numbers.



A sequence of rational numbers $(q_n)$ is said to be Cauchy if, for every positive rational $varepsilon$, there exists $k$ such that, for all $m,n>k$, $|q_m-q_n|<varepsilon$.



A sequence $(q_n)$ is a zero-sequence if, for every positive rational $varepsilon$, there exists $k$ such that, for all $n>k$, $|q_n|<varepsilon$. Every zero-sequence is Cauchy.



Fact 1. The set $mathscr{C}$ of Cauchy rational sequences forms a commutative ring under the operations $(q_n)+(r_n)=(q_n+r_n)$ and $(q_n)(r_n)=(q_nr_n)$.



The proof of this consists in showing that $mathscr{C}$ is a subring of the ring of all rational sequences with componentwise addition and multiplication.



Fact 2. The set $mathscr{Z}$ of zero-sequences is a maximal ideal of $mathscr{C}$.



The proof is direct verification.



Fact 3. The field $mathbb{Q}$ of rational numbers can be embedded in the quotient ring $mathbb{R}=mathscr{C}/mathscr{Z}$, which is a field as well.



The embedding consists in associating to a rational number the corresponding constant (Cauchy) sequence and verifying that in this way we get a ring homomorphism.



If $q=(q_n)$ is a Cauchy rational sequence, I'll denote with $hat{q}$ its image in the quotient ring $mathbb{R}$.



Fact 4. The field $mathbb{R}$ can be ordered.



Define $hat{q}<hat{r}$ if (and only if) there exists $k$ such that, for all $n>k$, $q_n<r_n$. It is readily seen that this is not dependent on the particular representative chosen for $hat{q}$ and $hat{r}$. Also the properties of ordered field are easy to verify.



Fact 5. (Cantor-Dedekind property) Every nonempty upper bounded subset of $mathbb{R}$ has a supremum.



This is the trickiest part. A closed interval is a subset of $mathbb{R}$ of the form $[hat{q},hat{r}]={hat{p}inmathbb{R}:hat{q}lehat{p}lehat{r}}$ (where $hat{p}<hat{r}$). A nested chain of closed intervals is given by two sequences $(hat{p}_n)$ and $(hat{r}_n)$ such that




  1. for all $n$, $hat{p}_n<hat{r}_n$;

  2. for all $m,n$, if $mle n$ then $hat{p}_mlehat{p}_n$ and $hat{r}_mgehat{r}_n$;

  3. for every positive rational $varepsilon$, there exists $k$ for which $hat{r}_k-hat{p}_k<varepsilon$.


It is not difficult (albeit tedious) to prove that, denoting by $pi_n$ the $n$-th term of a rational sequence representing $hat{p}_n$, and similarly defining $rho_n$ with the upper bounds, then





  1. $pi=(pi_n)$ and $rho=(rho_n)$ are Cauchy rational sequences;


  2. $hat{pi}=hat{rho}$;


  3. $hat{pi}=hat{rho}in[hat{p}_n,hat{r}_n]$, for every $n$.


In simpler words, given a nested chain of closed intervals, there exists a (necessarily unique) element of $mathbb{R}$ that belongs to every interval in the chain.



Deriving from this the Cantor-Dedekind property is standard.





As you see, the above construction of $mathbb{R}$ never uses the concept of limit, but it is of course modeled on it. Now it is essentially trivial to show that a Cauchy sequence in $mathbb{R}$ is convergent (with the standard definitions).



The construction of $mathbb{R}$ with Dedekind cuts allows for an easier derivation of the Cantor-Dedekind property. On the other hand, defining the field operations and verifying that we indeed get a field is much more boring. With the Cantor construction the field operations come almost for free from basic ring theory, but we pay a price when showing the Cantor-Dedekind property.



Both constructions give “the same” object, because it can be shown that two ordered fields having the Cantor-Dedekind property are isomorphic. The order relation on such a field is unique (essentially because every positive element has a square root). Such a field also contains a copy of the rational field $mathbb{Q}$ and is Archimedean.



Thus it is not relevant how we build the real numbers. Once we show that there exists a construction of an ordered field satisfying the Cantor-Dedekind property, we can free ourselves from the details of the construction and only use the properties of this field.



Now defining metric spaces with a metric taking its values in $mathbb{R}$ is not a problem any longer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:19












  • $begingroup$
    @alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
    $endgroup$
    – egreg
    Jan 7 at 15:21










  • $begingroup$
    dumb number used for irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:29










  • $begingroup$
    @alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
    $endgroup$
    – egreg
    Jan 7 at 15:47
















1












$begingroup$

Here's, in modern language, how Cantor defined the real numbers.



A sequence of rational numbers $(q_n)$ is said to be Cauchy if, for every positive rational $varepsilon$, there exists $k$ such that, for all $m,n>k$, $|q_m-q_n|<varepsilon$.



A sequence $(q_n)$ is a zero-sequence if, for every positive rational $varepsilon$, there exists $k$ such that, for all $n>k$, $|q_n|<varepsilon$. Every zero-sequence is Cauchy.



Fact 1. The set $mathscr{C}$ of Cauchy rational sequences forms a commutative ring under the operations $(q_n)+(r_n)=(q_n+r_n)$ and $(q_n)(r_n)=(q_nr_n)$.



The proof of this consists in showing that $mathscr{C}$ is a subring of the ring of all rational sequences with componentwise addition and multiplication.



Fact 2. The set $mathscr{Z}$ of zero-sequences is a maximal ideal of $mathscr{C}$.



The proof is direct verification.



Fact 3. The field $mathbb{Q}$ of rational numbers can be embedded in the quotient ring $mathbb{R}=mathscr{C}/mathscr{Z}$, which is a field as well.



The embedding consists in associating to a rational number the corresponding constant (Cauchy) sequence and verifying that in this way we get a ring homomorphism.



If $q=(q_n)$ is a Cauchy rational sequence, I'll denote with $hat{q}$ its image in the quotient ring $mathbb{R}$.



Fact 4. The field $mathbb{R}$ can be ordered.



Define $hat{q}<hat{r}$ if (and only if) there exists $k$ such that, for all $n>k$, $q_n<r_n$. It is readily seen that this is not dependent on the particular representative chosen for $hat{q}$ and $hat{r}$. Also the properties of ordered field are easy to verify.



Fact 5. (Cantor-Dedekind property) Every nonempty upper bounded subset of $mathbb{R}$ has a supremum.



This is the trickiest part. A closed interval is a subset of $mathbb{R}$ of the form $[hat{q},hat{r}]={hat{p}inmathbb{R}:hat{q}lehat{p}lehat{r}}$ (where $hat{p}<hat{r}$). A nested chain of closed intervals is given by two sequences $(hat{p}_n)$ and $(hat{r}_n)$ such that




  1. for all $n$, $hat{p}_n<hat{r}_n$;

  2. for all $m,n$, if $mle n$ then $hat{p}_mlehat{p}_n$ and $hat{r}_mgehat{r}_n$;

  3. for every positive rational $varepsilon$, there exists $k$ for which $hat{r}_k-hat{p}_k<varepsilon$.


It is not difficult (albeit tedious) to prove that, denoting by $pi_n$ the $n$-th term of a rational sequence representing $hat{p}_n$, and similarly defining $rho_n$ with the upper bounds, then





  1. $pi=(pi_n)$ and $rho=(rho_n)$ are Cauchy rational sequences;


  2. $hat{pi}=hat{rho}$;


  3. $hat{pi}=hat{rho}in[hat{p}_n,hat{r}_n]$, for every $n$.


In simpler words, given a nested chain of closed intervals, there exists a (necessarily unique) element of $mathbb{R}$ that belongs to every interval in the chain.



Deriving from this the Cantor-Dedekind property is standard.





As you see, the above construction of $mathbb{R}$ never uses the concept of limit, but it is of course modeled on it. Now it is essentially trivial to show that a Cauchy sequence in $mathbb{R}$ is convergent (with the standard definitions).



The construction of $mathbb{R}$ with Dedekind cuts allows for an easier derivation of the Cantor-Dedekind property. On the other hand, defining the field operations and verifying that we indeed get a field is much more boring. With the Cantor construction the field operations come almost for free from basic ring theory, but we pay a price when showing the Cantor-Dedekind property.



Both constructions give “the same” object, because it can be shown that two ordered fields having the Cantor-Dedekind property are isomorphic. The order relation on such a field is unique (essentially because every positive element has a square root). Such a field also contains a copy of the rational field $mathbb{Q}$ and is Archimedean.



Thus it is not relevant how we build the real numbers. Once we show that there exists a construction of an ordered field satisfying the Cantor-Dedekind property, we can free ourselves from the details of the construction and only use the properties of this field.



Now defining metric spaces with a metric taking its values in $mathbb{R}$ is not a problem any longer.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:19












  • $begingroup$
    @alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
    $endgroup$
    – egreg
    Jan 7 at 15:21










  • $begingroup$
    dumb number used for irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:29










  • $begingroup$
    @alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
    $endgroup$
    – egreg
    Jan 7 at 15:47














1












1








1





$begingroup$

Here's, in modern language, how Cantor defined the real numbers.



A sequence of rational numbers $(q_n)$ is said to be Cauchy if, for every positive rational $varepsilon$, there exists $k$ such that, for all $m,n>k$, $|q_m-q_n|<varepsilon$.



A sequence $(q_n)$ is a zero-sequence if, for every positive rational $varepsilon$, there exists $k$ such that, for all $n>k$, $|q_n|<varepsilon$. Every zero-sequence is Cauchy.



Fact 1. The set $mathscr{C}$ of Cauchy rational sequences forms a commutative ring under the operations $(q_n)+(r_n)=(q_n+r_n)$ and $(q_n)(r_n)=(q_nr_n)$.



The proof of this consists in showing that $mathscr{C}$ is a subring of the ring of all rational sequences with componentwise addition and multiplication.



Fact 2. The set $mathscr{Z}$ of zero-sequences is a maximal ideal of $mathscr{C}$.



The proof is direct verification.



Fact 3. The field $mathbb{Q}$ of rational numbers can be embedded in the quotient ring $mathbb{R}=mathscr{C}/mathscr{Z}$, which is a field as well.



The embedding consists in associating to a rational number the corresponding constant (Cauchy) sequence and verifying that in this way we get a ring homomorphism.



If $q=(q_n)$ is a Cauchy rational sequence, I'll denote with $hat{q}$ its image in the quotient ring $mathbb{R}$.



Fact 4. The field $mathbb{R}$ can be ordered.



Define $hat{q}<hat{r}$ if (and only if) there exists $k$ such that, for all $n>k$, $q_n<r_n$. It is readily seen that this is not dependent on the particular representative chosen for $hat{q}$ and $hat{r}$. Also the properties of ordered field are easy to verify.



Fact 5. (Cantor-Dedekind property) Every nonempty upper bounded subset of $mathbb{R}$ has a supremum.



This is the trickiest part. A closed interval is a subset of $mathbb{R}$ of the form $[hat{q},hat{r}]={hat{p}inmathbb{R}:hat{q}lehat{p}lehat{r}}$ (where $hat{p}<hat{r}$). A nested chain of closed intervals is given by two sequences $(hat{p}_n)$ and $(hat{r}_n)$ such that




  1. for all $n$, $hat{p}_n<hat{r}_n$;

  2. for all $m,n$, if $mle n$ then $hat{p}_mlehat{p}_n$ and $hat{r}_mgehat{r}_n$;

  3. for every positive rational $varepsilon$, there exists $k$ for which $hat{r}_k-hat{p}_k<varepsilon$.


It is not difficult (albeit tedious) to prove that, denoting by $pi_n$ the $n$-th term of a rational sequence representing $hat{p}_n$, and similarly defining $rho_n$ with the upper bounds, then





  1. $pi=(pi_n)$ and $rho=(rho_n)$ are Cauchy rational sequences;


  2. $hat{pi}=hat{rho}$;


  3. $hat{pi}=hat{rho}in[hat{p}_n,hat{r}_n]$, for every $n$.


In simpler words, given a nested chain of closed intervals, there exists a (necessarily unique) element of $mathbb{R}$ that belongs to every interval in the chain.



Deriving from this the Cantor-Dedekind property is standard.





As you see, the above construction of $mathbb{R}$ never uses the concept of limit, but it is of course modeled on it. Now it is essentially trivial to show that a Cauchy sequence in $mathbb{R}$ is convergent (with the standard definitions).



The construction of $mathbb{R}$ with Dedekind cuts allows for an easier derivation of the Cantor-Dedekind property. On the other hand, defining the field operations and verifying that we indeed get a field is much more boring. With the Cantor construction the field operations come almost for free from basic ring theory, but we pay a price when showing the Cantor-Dedekind property.



Both constructions give “the same” object, because it can be shown that two ordered fields having the Cantor-Dedekind property are isomorphic. The order relation on such a field is unique (essentially because every positive element has a square root). Such a field also contains a copy of the rational field $mathbb{Q}$ and is Archimedean.



Thus it is not relevant how we build the real numbers. Once we show that there exists a construction of an ordered field satisfying the Cantor-Dedekind property, we can free ourselves from the details of the construction and only use the properties of this field.



Now defining metric spaces with a metric taking its values in $mathbb{R}$ is not a problem any longer.






share|cite|improve this answer









$endgroup$



Here's, in modern language, how Cantor defined the real numbers.



A sequence of rational numbers $(q_n)$ is said to be Cauchy if, for every positive rational $varepsilon$, there exists $k$ such that, for all $m,n>k$, $|q_m-q_n|<varepsilon$.



A sequence $(q_n)$ is a zero-sequence if, for every positive rational $varepsilon$, there exists $k$ such that, for all $n>k$, $|q_n|<varepsilon$. Every zero-sequence is Cauchy.



Fact 1. The set $mathscr{C}$ of Cauchy rational sequences forms a commutative ring under the operations $(q_n)+(r_n)=(q_n+r_n)$ and $(q_n)(r_n)=(q_nr_n)$.



The proof of this consists in showing that $mathscr{C}$ is a subring of the ring of all rational sequences with componentwise addition and multiplication.



Fact 2. The set $mathscr{Z}$ of zero-sequences is a maximal ideal of $mathscr{C}$.



The proof is direct verification.



Fact 3. The field $mathbb{Q}$ of rational numbers can be embedded in the quotient ring $mathbb{R}=mathscr{C}/mathscr{Z}$, which is a field as well.



The embedding consists in associating to a rational number the corresponding constant (Cauchy) sequence and verifying that in this way we get a ring homomorphism.



If $q=(q_n)$ is a Cauchy rational sequence, I'll denote with $hat{q}$ its image in the quotient ring $mathbb{R}$.



Fact 4. The field $mathbb{R}$ can be ordered.



Define $hat{q}<hat{r}$ if (and only if) there exists $k$ such that, for all $n>k$, $q_n<r_n$. It is readily seen that this is not dependent on the particular representative chosen for $hat{q}$ and $hat{r}$. Also the properties of ordered field are easy to verify.



Fact 5. (Cantor-Dedekind property) Every nonempty upper bounded subset of $mathbb{R}$ has a supremum.



This is the trickiest part. A closed interval is a subset of $mathbb{R}$ of the form $[hat{q},hat{r}]={hat{p}inmathbb{R}:hat{q}lehat{p}lehat{r}}$ (where $hat{p}<hat{r}$). A nested chain of closed intervals is given by two sequences $(hat{p}_n)$ and $(hat{r}_n)$ such that




  1. for all $n$, $hat{p}_n<hat{r}_n$;

  2. for all $m,n$, if $mle n$ then $hat{p}_mlehat{p}_n$ and $hat{r}_mgehat{r}_n$;

  3. for every positive rational $varepsilon$, there exists $k$ for which $hat{r}_k-hat{p}_k<varepsilon$.


It is not difficult (albeit tedious) to prove that, denoting by $pi_n$ the $n$-th term of a rational sequence representing $hat{p}_n$, and similarly defining $rho_n$ with the upper bounds, then





  1. $pi=(pi_n)$ and $rho=(rho_n)$ are Cauchy rational sequences;


  2. $hat{pi}=hat{rho}$;


  3. $hat{pi}=hat{rho}in[hat{p}_n,hat{r}_n]$, for every $n$.


In simpler words, given a nested chain of closed intervals, there exists a (necessarily unique) element of $mathbb{R}$ that belongs to every interval in the chain.



Deriving from this the Cantor-Dedekind property is standard.





As you see, the above construction of $mathbb{R}$ never uses the concept of limit, but it is of course modeled on it. Now it is essentially trivial to show that a Cauchy sequence in $mathbb{R}$ is convergent (with the standard definitions).



The construction of $mathbb{R}$ with Dedekind cuts allows for an easier derivation of the Cantor-Dedekind property. On the other hand, defining the field operations and verifying that we indeed get a field is much more boring. With the Cantor construction the field operations come almost for free from basic ring theory, but we pay a price when showing the Cantor-Dedekind property.



Both constructions give “the same” object, because it can be shown that two ordered fields having the Cantor-Dedekind property are isomorphic. The order relation on such a field is unique (essentially because every positive element has a square root). Such a field also contains a copy of the rational field $mathbb{Q}$ and is Archimedean.



Thus it is not relevant how we build the real numbers. Once we show that there exists a construction of an ordered field satisfying the Cantor-Dedekind property, we can free ourselves from the details of the construction and only use the properties of this field.



Now defining metric spaces with a metric taking its values in $mathbb{R}$ is not a problem any longer.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 22:24









egregegreg

185k1486206




185k1486206












  • $begingroup$
    In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:19












  • $begingroup$
    @alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
    $endgroup$
    – egreg
    Jan 7 at 15:21










  • $begingroup$
    dumb number used for irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:29










  • $begingroup$
    @alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
    $endgroup$
    – egreg
    Jan 7 at 15:47


















  • $begingroup$
    In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
    $endgroup$
    – alwaystudent
    Jan 7 at 15:19












  • $begingroup$
    @alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
    $endgroup$
    – egreg
    Jan 7 at 15:21










  • $begingroup$
    dumb number used for irrational number
    $endgroup$
    – alwaystudent
    Jan 7 at 15:29










  • $begingroup$
    @alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
    $endgroup$
    – egreg
    Jan 7 at 15:47
















$begingroup$
In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
$endgroup$
– alwaystudent
Jan 7 at 15:19






$begingroup$
In fact, I do not remember my original proof , but can not say that Q in Q-valued metric space is complete and And there's no place for dumb numbers?
$endgroup$
– alwaystudent
Jan 7 at 15:19














$begingroup$
@alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
$endgroup$
– egreg
Jan 7 at 15:21




$begingroup$
@alwaystudent What's a dumb number? I don't need $mathbb{Q}$-valued metric spaces at any place.
$endgroup$
– egreg
Jan 7 at 15:21












$begingroup$
dumb number used for irrational number
$endgroup$
– alwaystudent
Jan 7 at 15:29




$begingroup$
dumb number used for irrational number
$endgroup$
– alwaystudent
Jan 7 at 15:29












$begingroup$
@alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
$endgroup$
– egreg
Jan 7 at 15:47




$begingroup$
@alwaystudent I can't see the problem. I never defined irrational numbers. They just arise from the definition as the real numbers that happen to not be rational.
$endgroup$
– egreg
Jan 7 at 15:47











0












$begingroup$

One way of defining the real numbers is to say that they form an ordered field in which every non-empty subset which is bounded above has a least upper bound.



That then begs the question of whether such a field actually exists. The construction using Dedekind cuts, or alternatively using Cauchy sequences of the rationals, is sufficient to prove existence, and in fact then the least upper bound property does what you need.



There is also the question of whether such a field is unique, and indeed it is.



It is a common feature of mathematical definitions that a property is defined, and then is shown to be equivalent to some other useful properties and also to imply others. There is work to be done whichever equivalent definition is chosen, and sometimes we choose the definition which is easiest to work with in the context we have in mind.





So it depends a bit on whether you are using limits to define the real numbers, or to construct them, and then show that the numbers you construct have the properties you need.



If you define the real numbers via Cauchy Sequences of the Rationals, then there is a logical step to show that Cauchy Sequences of the Reals behave as expected (as Henning Makholm has indicated). Whichever way you do it some care has to be taken - it is not as easy as it might at first appear to show that real numbers constructed as Dedekind Sections obey the usual rules of arithmetic. However, the details have been worked through, and efficient proofs are known. But usually much of this detail is taken for granted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
    $endgroup$
    – Mark Bennet
    Jan 6 at 21:14
















0












$begingroup$

One way of defining the real numbers is to say that they form an ordered field in which every non-empty subset which is bounded above has a least upper bound.



That then begs the question of whether such a field actually exists. The construction using Dedekind cuts, or alternatively using Cauchy sequences of the rationals, is sufficient to prove existence, and in fact then the least upper bound property does what you need.



There is also the question of whether such a field is unique, and indeed it is.



It is a common feature of mathematical definitions that a property is defined, and then is shown to be equivalent to some other useful properties and also to imply others. There is work to be done whichever equivalent definition is chosen, and sometimes we choose the definition which is easiest to work with in the context we have in mind.





So it depends a bit on whether you are using limits to define the real numbers, or to construct them, and then show that the numbers you construct have the properties you need.



If you define the real numbers via Cauchy Sequences of the Rationals, then there is a logical step to show that Cauchy Sequences of the Reals behave as expected (as Henning Makholm has indicated). Whichever way you do it some care has to be taken - it is not as easy as it might at first appear to show that real numbers constructed as Dedekind Sections obey the usual rules of arithmetic. However, the details have been worked through, and efficient proofs are known. But usually much of this detail is taken for granted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
    $endgroup$
    – Mark Bennet
    Jan 6 at 21:14














0












0








0





$begingroup$

One way of defining the real numbers is to say that they form an ordered field in which every non-empty subset which is bounded above has a least upper bound.



That then begs the question of whether such a field actually exists. The construction using Dedekind cuts, or alternatively using Cauchy sequences of the rationals, is sufficient to prove existence, and in fact then the least upper bound property does what you need.



There is also the question of whether such a field is unique, and indeed it is.



It is a common feature of mathematical definitions that a property is defined, and then is shown to be equivalent to some other useful properties and also to imply others. There is work to be done whichever equivalent definition is chosen, and sometimes we choose the definition which is easiest to work with in the context we have in mind.





So it depends a bit on whether you are using limits to define the real numbers, or to construct them, and then show that the numbers you construct have the properties you need.



If you define the real numbers via Cauchy Sequences of the Rationals, then there is a logical step to show that Cauchy Sequences of the Reals behave as expected (as Henning Makholm has indicated). Whichever way you do it some care has to be taken - it is not as easy as it might at first appear to show that real numbers constructed as Dedekind Sections obey the usual rules of arithmetic. However, the details have been worked through, and efficient proofs are known. But usually much of this detail is taken for granted.






share|cite|improve this answer











$endgroup$



One way of defining the real numbers is to say that they form an ordered field in which every non-empty subset which is bounded above has a least upper bound.



That then begs the question of whether such a field actually exists. The construction using Dedekind cuts, or alternatively using Cauchy sequences of the rationals, is sufficient to prove existence, and in fact then the least upper bound property does what you need.



There is also the question of whether such a field is unique, and indeed it is.



It is a common feature of mathematical definitions that a property is defined, and then is shown to be equivalent to some other useful properties and also to imply others. There is work to be done whichever equivalent definition is chosen, and sometimes we choose the definition which is easiest to work with in the context we have in mind.





So it depends a bit on whether you are using limits to define the real numbers, or to construct them, and then show that the numbers you construct have the properties you need.



If you define the real numbers via Cauchy Sequences of the Rationals, then there is a logical step to show that Cauchy Sequences of the Reals behave as expected (as Henning Makholm has indicated). Whichever way you do it some care has to be taken - it is not as easy as it might at first appear to show that real numbers constructed as Dedekind Sections obey the usual rules of arithmetic. However, the details have been worked through, and efficient proofs are known. But usually much of this detail is taken for granted.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 6 at 21:08

























answered Jan 6 at 20:53









Mark BennetMark Bennet

81.7k984182




81.7k984182












  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
    $endgroup$
    – Mark Bennet
    Jan 6 at 21:14


















  • $begingroup$
    I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
    $endgroup$
    – alwaystudent
    Jan 6 at 21:05








  • 1




    $begingroup$
    @alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
    $endgroup$
    – Mark Bennet
    Jan 6 at 21:14
















$begingroup$
I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
$endgroup$
– alwaystudent
Jan 6 at 21:05






$begingroup$
I already looked at it and it became clear that the rational numbers for the metric space are not sufficient and dedkind cut is too problematic too,In fact, real numbers are the concept of geometry
$endgroup$
– alwaystudent
Jan 6 at 21:05






1




1




$begingroup$
@alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
$endgroup$
– Mark Bennet
Jan 6 at 21:14




$begingroup$
@alwaystudent Obviously the rationals will not do if you want limits. If you want every Cauchy Sequence to have a limit, you have to have a way of giving a limit to every such sequence, so one form or another of the construction becomes almost inevitable and you have to make it work. The difference, in the end, is the way in which different sequences which have the same limit are treated. In the Cauchy method they are regarded as differing by a null sequence. In the Dedekind method they are seen as defining the same section of the rationals (and later the reals).
$endgroup$
– Mark Bennet
Jan 6 at 21:14


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064360%2freal-numbers-definition-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Bressuire

Cabo Verde

Gyllenstierna