Trouble proving that $bigcup mathbb N = mathbb N$
$begingroup$
Prove that $bigcup mathbb N = mathbb N$
I have no idea how to even approach this question; I know that $mathbb N$ is defined as the intersection of all inductive subset of any inductive set $y$, and intersection is related to union in the sense that $bigcap x subseteq bigcup x$, where $x$ is any non-empty set - but that is all I can think of.
This question is an exercise of a chapter related to ZF 7-9 (ie. ax. of infinity, replacement and foundation), so I suspect I will have to utilise ax. of infinity at some point, but I just don't know where it fits in.
Ax. of infinity guarantees that $mathbb N$ is a set; and ax. of union also guarantees that $cup mathbb N$ is a set - but I don't suppose that these are at all helpful.
Could anyone help or at least give me a sense of direction please?
set-theory proof-explanation
edited Dec 23 '18 at 15:47
egreg
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
$endgroup$
add a comment |
$begingroup$
Prove that $bigcup mathbb N = mathbb N$
I have no idea how to even approach this question; I know that $mathbb N$ is defined as the intersection of all inductive subset of any inductive set $y$, and intersection is related to union in the sense that $bigcap x subseteq bigcup x$, where $x$ is any non-empty set - but that is all I can think of.
This question is an exercise of a chapter related to ZF 7-9 (ie. ax. of infinity, replacement and foundation), so I suspect I will have to utilise ax. of infinity at some point, but I just don't know where it fits in.
Ax. of infinity guarantees that $mathbb N$ is a set; and ax. of union also guarantees that $cup mathbb N$ is a set - but I don't suppose that these are at all helpful.
Could anyone help or at least give me a sense of direction please?
set-theory proof-explanation
edited Dec 23 '18 at 15:47
egreg
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
$endgroup$
1
$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
1
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45
add a comment |
$begingroup$
Prove that $bigcup mathbb N = mathbb N$
I have no idea how to even approach this question; I know that $mathbb N$ is defined as the intersection of all inductive subset of any inductive set $y$, and intersection is related to union in the sense that $bigcap x subseteq bigcup x$, where $x$ is any non-empty set - but that is all I can think of.
This question is an exercise of a chapter related to ZF 7-9 (ie. ax. of infinity, replacement and foundation), so I suspect I will have to utilise ax. of infinity at some point, but I just don't know where it fits in.
Ax. of infinity guarantees that $mathbb N$ is a set; and ax. of union also guarantees that $cup mathbb N$ is a set - but I don't suppose that these are at all helpful.
Could anyone help or at least give me a sense of direction please?
set-theory proof-explanation
edited Dec 23 '18 at 15:47
egreg
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
$endgroup$
Prove that $bigcup mathbb N = mathbb N$
I have no idea how to even approach this question; I know that $mathbb N$ is defined as the intersection of all inductive subset of any inductive set $y$, and intersection is related to union in the sense that $bigcap x subseteq bigcup x$, where $x$ is any non-empty set - but that is all I can think of.
This question is an exercise of a chapter related to ZF 7-9 (ie. ax. of infinity, replacement and foundation), so I suspect I will have to utilise ax. of infinity at some point, but I just don't know where it fits in.
Ax. of infinity guarantees that $mathbb N$ is a set; and ax. of union also guarantees that $cup mathbb N$ is a set - but I don't suppose that these are at all helpful.
Could anyone help or at least give me a sense of direction please?
set-theory proof-explanation
set-theory proof-explanation
edited Dec 23 '18 at 15:47
egreg
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
edited Dec 23 '18 at 15:47
egreg
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
edited Dec 23 '18 at 15:47
egreg
181k1485203
edited Dec 23 '18 at 15:47
egreg
181k1485203
edited Dec 23 '18 at 15:47
egreg
181k1485203
181k1485203
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
asked Dec 23 '18 at 15:28
Daniel MakDaniel Mak
447416
447416
1
$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
1
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45
add a comment |
1
$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
1
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45
1
1
$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
1
1
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It depends on how you define $mathbb{N}$ to begin with. If you define it as the least inductive set containing $emptyset$ as an element, then $mathbb{N}$ is a (von Neumann) ordinal, hence it is $in$-transitive: if $ain b$ and $bin mathbb{N}$, then $ainmathbb{N}$. This proves $bigcupmathbb{N}subseteqmathbb{N}$.
Further, if $ainmathbb{N}$, then $ain a+1$, so $mathbb{N}subseteqbigcupmathbb{N}$.
By the way, this is also true for every limit ordinal. If $alpha=beta+1$ is a successor ordinal, then $bigcupalpha=beta$.
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".
Once you have that, you can do a very straightforward proof that the union of all of the elements of $mathbb{N}$ is itself $mathbb{N}$. For instance, pick an element of $mathbb{N}$, say $5$. Well, $5in6inmathbb{N}$, so $5incupmathbb{N}$.
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
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add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
votes
$begingroup$
It depends on how you define $mathbb{N}$ to begin with. If you define it as the least inductive set containing $emptyset$ as an element, then $mathbb{N}$ is a (von Neumann) ordinal, hence it is $in$-transitive: if $ain b$ and $bin mathbb{N}$, then $ainmathbb{N}$. This proves $bigcupmathbb{N}subseteqmathbb{N}$.
Further, if $ainmathbb{N}$, then $ain a+1$, so $mathbb{N}subseteqbigcupmathbb{N}$.
By the way, this is also true for every limit ordinal. If $alpha=beta+1$ is a successor ordinal, then $bigcupalpha=beta$.
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
It depends on how you define $mathbb{N}$ to begin with. If you define it as the least inductive set containing $emptyset$ as an element, then $mathbb{N}$ is a (von Neumann) ordinal, hence it is $in$-transitive: if $ain b$ and $bin mathbb{N}$, then $ainmathbb{N}$. This proves $bigcupmathbb{N}subseteqmathbb{N}$.
Further, if $ainmathbb{N}$, then $ain a+1$, so $mathbb{N}subseteqbigcupmathbb{N}$.
By the way, this is also true for every limit ordinal. If $alpha=beta+1$ is a successor ordinal, then $bigcupalpha=beta$.
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
$endgroup$
add a comment |
$begingroup$
It depends on how you define $mathbb{N}$ to begin with. If you define it as the least inductive set containing $emptyset$ as an element, then $mathbb{N}$ is a (von Neumann) ordinal, hence it is $in$-transitive: if $ain b$ and $bin mathbb{N}$, then $ainmathbb{N}$. This proves $bigcupmathbb{N}subseteqmathbb{N}$.
Further, if $ainmathbb{N}$, then $ain a+1$, so $mathbb{N}subseteqbigcupmathbb{N}$.
By the way, this is also true for every limit ordinal. If $alpha=beta+1$ is a successor ordinal, then $bigcupalpha=beta$.
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
$endgroup$
It depends on how you define $mathbb{N}$ to begin with. If you define it as the least inductive set containing $emptyset$ as an element, then $mathbb{N}$ is a (von Neumann) ordinal, hence it is $in$-transitive: if $ain b$ and $bin mathbb{N}$, then $ainmathbb{N}$. This proves $bigcupmathbb{N}subseteqmathbb{N}$.
Further, if $ainmathbb{N}$, then $ain a+1$, so $mathbb{N}subseteqbigcupmathbb{N}$.
By the way, this is also true for every limit ordinal. If $alpha=beta+1$ is a successor ordinal, then $bigcupalpha=beta$.
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
answered Dec 23 '18 at 15:54
egregegreg
181k1485203
181k1485203
add a comment |
add a comment |
$begingroup$
I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".
Once you have that, you can do a very straightforward proof that the union of all of the elements of $mathbb{N}$ is itself $mathbb{N}$. For instance, pick an element of $mathbb{N}$, say $5$. Well, $5in6inmathbb{N}$, so $5incupmathbb{N}$.
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".
Once you have that, you can do a very straightforward proof that the union of all of the elements of $mathbb{N}$ is itself $mathbb{N}$. For instance, pick an element of $mathbb{N}$, say $5$. Well, $5in6inmathbb{N}$, so $5incupmathbb{N}$.
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
$endgroup$
add a comment |
$begingroup$
I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".
Once you have that, you can do a very straightforward proof that the union of all of the elements of $mathbb{N}$ is itself $mathbb{N}$. For instance, pick an element of $mathbb{N}$, say $5$. Well, $5in6inmathbb{N}$, so $5incupmathbb{N}$.
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
$endgroup$
I think the issue will resolve itself if you go back to some definitions; taking from the wikipedia article on the axiom of infinity, you can have a very explicit set theoretic definition of the elements of $mathbb{N}$ beyond the usual "intersection of all inductive subsets of any inductive set".
Once you have that, you can do a very straightforward proof that the union of all of the elements of $mathbb{N}$ is itself $mathbb{N}$. For instance, pick an element of $mathbb{N}$, say $5$. Well, $5in6inmathbb{N}$, so $5incupmathbb{N}$.
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
answered Dec 23 '18 at 15:42
RandomNumberGuyRandomNumberGuy
662
662
add a comment |
add a comment |
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$begingroup$
How do you define $cup mathbb N $?
$endgroup$
– Dietrich Burde
Dec 23 '18 at 15:38
1
$begingroup$
@Dietrich: First you define $bigcup X$, then you take $X=Bbb N$. Specifically, $bigcup X={ymidexists xin X: yin x}$, and now plug in $Bbb N$.
$endgroup$
– Asaf Karagila♦
Dec 23 '18 at 16:45