Prove that the set used in archimedean property is closed
$begingroup$
Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.
What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.
Is my approach right ?
Please help!
real-analysis
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
add a comment |
$begingroup$
Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.
What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.
Is my approach right ?
Please help!
real-analysis
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26
add a comment |
$begingroup$
Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.
What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.
Is my approach right ?
Please help!
real-analysis
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
$endgroup$
Let $x$ be a positive real number. Show that the set
${nx:ninBbb N}$ is closed in $Bbb R$.
What I tried to do is that consider the function $f (x)=nx$ which is continuous in $Bbb R$ and tried to show that $g (x)=f (x)-nx$ greater than $0$ or less than $0$ is an open set.
Is my approach right ?
Please help!
real-analysis
real-analysis
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
edited Dec 17 '18 at 4:12
Tianlalu
3,08621038
3,08621038
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
asked Dec 17 '18 at 4:07
Debarghya MukherjeeDebarghya Mukherjee
352
352
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26
add a comment |
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have a couple options:
1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.
Clearly the first case is easy. The second is slightly harder.
2) Show the set contains all its limit points.
This might be a bit harder.
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
$endgroup$
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
add a comment |
$begingroup$
(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)
And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.
I don't see this approach working.
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043511%2fprove-that-the-set-used-in-archimedean-property-is-closed%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have a couple options:
1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.
Clearly the first case is easy. The second is slightly harder.
2) Show the set contains all its limit points.
This might be a bit harder.
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
$endgroup$
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
add a comment |
$begingroup$
You have a couple options:
1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.
Clearly the first case is easy. The second is slightly harder.
2) Show the set contains all its limit points.
This might be a bit harder.
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
$endgroup$
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
add a comment |
$begingroup$
You have a couple options:
1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.
Clearly the first case is easy. The second is slightly harder.
2) Show the set contains all its limit points.
This might be a bit harder.
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
$endgroup$
You have a couple options:
1) Show the compliment is open.
That’s is, the set $mathbb{R}setminus mathbb{xN}$ is open. This is easy since every number in this set can reside in a ball of radius smaller than the nearest $nx$. This might need to be done using two separate cases: $x = 0$ and $x not = 0$.
Clearly the first case is easy. The second is slightly harder.
2) Show the set contains all its limit points.
This might be a bit harder.
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
answered Dec 17 '18 at 4:23
gdepaulgdepaul
613
613
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
add a comment |
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
So do we need to apply archime dean property
$endgroup$
– Debarghya Mukherjee
Dec 17 '18 at 4:33
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
For the case x is not zero, then we take a number in the set R N, say y. Now dividing that number by x, we get, y/x. By the Archimedean property, there exist n such that n <= y/x <= n+1. So xn <= y <= x(n+1). Now you just need to show that this y is in an open ball
$endgroup$
– gdepaul
Dec 17 '18 at 4:36
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
$begingroup$
Complement, not compliment. Related to the word "complete".
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:21
add a comment |
$begingroup$
(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)
And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.
I don't see this approach working.
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
$endgroup$
add a comment |
$begingroup$
(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)
And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.
I don't see this approach working.
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
$endgroup$
add a comment |
$begingroup$
(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)
And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.
I don't see this approach working.
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
$endgroup$
(gdepaul's answer contains suggestions for alternate approaches. This one focuses on the idea laid out in the question)
And where would you go from there? You went from a fixed $x$ and an infinite collection of $n$ to a fixed $n$ and variable $x$; to recover the set we care about, we would have to take an intersection of infinitely many versions. That - well, an intersection of infinitely many open sets is not necessarily open.
I don't see this approach working.
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
answered Dec 17 '18 at 4:24
jmerryjmerry
4,334514
4,334514
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043511%2fprove-that-the-set-used-in-archimedean-property-is-closed%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
I dk what you are thinking. If $f(x)=x$ then $0=nx-nx=f(x)-nx=g(x).$
$endgroup$
– DanielWainfleet
Dec 17 '18 at 6:26