Mixing
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Proof feedback - Proving infinite intersections of compact sets is compact.
$begingroup$
I am looking to prove that the infinite intersections of compact sets is compact. I'd like some feedback on my proof.
Note: This course isn't a formal course in topology and we specifically work in $mathbb{R}^n$. I have done some reading and there seems to be some discussion about this question in other spaces.
Proof.
Let $A_1$ and $A_2$ be arbitrary sets. We need to show that $A_1 cap A_2$ is compact.
Let $O$ be the open cover $A_1 cup A_2$ which completely covers $A_1 cap A_2$ since if $x in A_1 cap A_2$ then $x in A_1$ and $x in A_2$, which is completely covered by $O$.
Since $A_1$ and $A_2$ are compact, then there exists a finite subcover $O_1 subset O$ and $O_2 subset O$ respectively which cover the respective sets. So $O_1 cup O_2$ cover $A_1 cup A_2$. Therefore $O_1 cap O_2$ covers $A_1 cap A_2$. (Perhaps I can justify this better?)
Since the sets were arbitrary, we have that the infinite intersection is compact. QED.
Let me know what you think. Thanks!
general-topology
edited Oct 29 '16 at 23:05
AJY
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
$endgroup$
|
show 4 more comments
$begingroup$
I am looking to prove that the infinite intersections of compact sets is compact. I'd like some feedback on my proof.
Note: This course isn't a formal course in topology and we specifically work in $mathbb{R}^n$. I have done some reading and there seems to be some discussion about this question in other spaces.
Proof.
Let $A_1$ and $A_2$ be arbitrary sets. We need to show that $A_1 cap A_2$ is compact.
Let $O$ be the open cover $A_1 cup A_2$ which completely covers $A_1 cap A_2$ since if $x in A_1 cap A_2$ then $x in A_1$ and $x in A_2$, which is completely covered by $O$.
Since $A_1$ and $A_2$ are compact, then there exists a finite subcover $O_1 subset O$ and $O_2 subset O$ respectively which cover the respective sets. So $O_1 cup O_2$ cover $A_1 cup A_2$. Therefore $O_1 cap O_2$ covers $A_1 cap A_2$. (Perhaps I can justify this better?)
Since the sets were arbitrary, we have that the infinite intersection is compact. QED.
Let me know what you think. Thanks!
general-topology
edited Oct 29 '16 at 23:05
AJY
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
$endgroup$
1
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
1
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30
|
show 4 more comments
$begingroup$
I am looking to prove that the infinite intersections of compact sets is compact. I'd like some feedback on my proof.
Note: This course isn't a formal course in topology and we specifically work in $mathbb{R}^n$. I have done some reading and there seems to be some discussion about this question in other spaces.
Proof.
Let $A_1$ and $A_2$ be arbitrary sets. We need to show that $A_1 cap A_2$ is compact.
Let $O$ be the open cover $A_1 cup A_2$ which completely covers $A_1 cap A_2$ since if $x in A_1 cap A_2$ then $x in A_1$ and $x in A_2$, which is completely covered by $O$.
Since $A_1$ and $A_2$ are compact, then there exists a finite subcover $O_1 subset O$ and $O_2 subset O$ respectively which cover the respective sets. So $O_1 cup O_2$ cover $A_1 cup A_2$. Therefore $O_1 cap O_2$ covers $A_1 cap A_2$. (Perhaps I can justify this better?)
Since the sets were arbitrary, we have that the infinite intersection is compact. QED.
Let me know what you think. Thanks!
general-topology
edited Oct 29 '16 at 23:05
AJY
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
$endgroup$
I am looking to prove that the infinite intersections of compact sets is compact. I'd like some feedback on my proof.
Note: This course isn't a formal course in topology and we specifically work in $mathbb{R}^n$. I have done some reading and there seems to be some discussion about this question in other spaces.
Proof.
Let $A_1$ and $A_2$ be arbitrary sets. We need to show that $A_1 cap A_2$ is compact.
Let $O$ be the open cover $A_1 cup A_2$ which completely covers $A_1 cap A_2$ since if $x in A_1 cap A_2$ then $x in A_1$ and $x in A_2$, which is completely covered by $O$.
Since $A_1$ and $A_2$ are compact, then there exists a finite subcover $O_1 subset O$ and $O_2 subset O$ respectively which cover the respective sets. So $O_1 cup O_2$ cover $A_1 cup A_2$. Therefore $O_1 cap O_2$ covers $A_1 cap A_2$. (Perhaps I can justify this better?)
Since the sets were arbitrary, we have that the infinite intersection is compact. QED.
Let me know what you think. Thanks!
general-topology
general-topology
edited Oct 29 '16 at 23:05
AJY
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
edited Oct 29 '16 at 23:05
AJY
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
edited Oct 29 '16 at 23:05
AJY
4,27021129
edited Oct 29 '16 at 23:05
AJY
4,27021129
edited Oct 29 '16 at 23:05
AJY
4,27021129
4,27021129
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
asked Oct 29 '16 at 17:59
TimelordViktoriousTimelordViktorious
487313
487313
1
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
1
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30
|
show 4 more comments
1
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
1
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30
1
1
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
1
1
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1cap A_2$ instead of taking an open cover for $A_1cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1cup A_2$, say $O'=Ocup{mathbb{R}^nsetminus Acap B}$, because $A_1cap A_2$ is closed.
Also, when you claim that $O_1cap O_2$ covers $A_1cap A_2$, I believe we may get some problems with this. For if $xin A_1cap A_2$, can we always find a single $Uin O_1cap O_2$ such that $xin U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $Ksubsetmathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
$endgroup$
add a comment |
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$begingroup$
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1cap A_2$ instead of taking an open cover for $A_1cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1cup A_2$, say $O'=Ocup{mathbb{R}^nsetminus Acap B}$, because $A_1cap A_2$ is closed.
Also, when you claim that $O_1cap O_2$ covers $A_1cap A_2$, I believe we may get some problems with this. For if $xin A_1cap A_2$, can we always find a single $Uin O_1cap O_2$ such that $xin U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $Ksubsetmathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
$endgroup$
add a comment |
$begingroup$
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1cap A_2$ instead of taking an open cover for $A_1cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1cup A_2$, say $O'=Ocup{mathbb{R}^nsetminus Acap B}$, because $A_1cap A_2$ is closed.
Also, when you claim that $O_1cap O_2$ covers $A_1cap A_2$, I believe we may get some problems with this. For if $xin A_1cap A_2$, can we always find a single $Uin O_1cap O_2$ such that $xin U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $Ksubsetmathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
$endgroup$
add a comment |
$begingroup$
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1cap A_2$ instead of taking an open cover for $A_1cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1cup A_2$, say $O'=Ocup{mathbb{R}^nsetminus Acap B}$, because $A_1cap A_2$ is closed.
Also, when you claim that $O_1cap O_2$ covers $A_1cap A_2$, I believe we may get some problems with this. For if $xin A_1cap A_2$, can we always find a single $Uin O_1cap O_2$ such that $xin U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $Ksubsetmathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
$endgroup$
The following are my messages (with a few updates) in the chat with MH1993, I moved them to here because I'm not sure if he can access it.
Well, in my opinion, you should start with an open cover $O$ of $A_1cap A_2$ instead of taking an open cover for $A_1cup A_2$ from the beginning. Of course, you can induce an open cover for $A_1cup A_2$, say $O'=Ocup{mathbb{R}^nsetminus Acap B}$, because $A_1cap A_2$ is closed.
Also, when you claim that $O_1cap O_2$ covers $A_1cap A_2$, I believe we may get some problems with this. For if $xin A_1cap A_2$, can we always find a single $Uin O_1cap O_2$ such that $xin U$? It could be the case that none of the open sets in $O_1$ belong to $O_2$, vice-versa.
Finally, as far as I can see, by showing that $A_1cap A_2$ is compact you can conclude by induction that the intersection of finitely many compact sets is compact, but you can't conclude that the intersection of infinitely many compact sets is compact. For example, one cannot conclude that since "the product of any two compact sets is compact" then "the arbitrary product of compact sets is compact": the former is true in ZF while the later is equivalent to the axiom of choice. Maybe there is a way to prove what you want in the way you want, but I don't know how to do it.
However, in any case, you need to use the fact that if $Ksubsetmathbb{R}^n$ is compact, then $K$ is closed. Hence any intersection of compact sets is closed and bounded. So, Heine-Borel Theorem handles the arbitrary case without many efforts.
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
answered Oct 29 '16 at 22:55
Renan Maneli MezabarbaRenan Maneli Mezabarba
1,000615
1,000615
add a comment |
add a comment |
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1
$begingroup$
It seems a little confusing to me the use of $A_1cup A_2$... Anyway, since you are working in $mathbb{R}^n$, notice that it is a Hausdorff space. Compact subspaces of Hausdorff spaces are also closed, hence the arbitrary intersection of compact sets is closed. Now, in general, closed subspaces of compact spaces are compact.
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:22
$begingroup$
I can't use anything about Hausdorff spaces.
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:23
1
$begingroup$
Can you use Heine-Borel theorem ($Ksubset mathbb{R}^n$ is compact if and only if $K$ is closed and bounded)?
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:26
$begingroup$
Yep. I could show that the sets are closed/bounded in separate cases and apply Heine-Borel theorem. Buuut I did want to try out with open covers and finite subcovers first. :p
$endgroup$
– TimelordViktorious
Oct 29 '16 at 18:28
$begingroup$
Maybe this math.stackexchange.com/questions/1499402/… will be useful
$endgroup$
– Renan Maneli Mezabarba
Oct 29 '16 at 18:30