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Does there exist a closed set which is an intersection of a collection of infinite open sets? [on hold]

Analysis Open and Closed SetsIf $F_1$ and $F_2$ are disjoint closed sets then there exist disjoint open sets $G_1$ and $G_2$.Intersection of finite collection of open sets is openA set that is a countable intersection of open and dense sets but not open.Showing the intersection of infinitely many open sets is closedInfinite intersection of open sets.Union of infinite open and close setsShowing that if closed subsets don't intersect then there exists open sets in which they exist that also don't intersectX compact iff any collection of closed set closed by finite intersection have a nonempty intersectionIs there an analogue for a compact set using closed sets?

0

$begingroup$

Does there exist a closed set which is an intersection of a collection of infinite open sets?

analysis

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edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

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Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Eevee Trainer, Shalop, Shailesh, user21820, José Carlos Santos Mar 18 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Shalop, Shailesh, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Intersect $(-tfrac1n, tfrac1n)$ for $n = 1, 2, ldots$ and consider what set you get
  $endgroup$
  – Brevan Ellefsen
  Mar 18 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh it will get $0$
  $endgroup$
  – Tony Tong
  Mar 18 at 5:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrevanEllefsen: The question only asks for a "collection of infinite open sets", so just the singleton $(0,1)$ suffices. But Tony, please read how to ask a good question.
  $endgroup$
  – user21820
  Mar 18 at 9:23
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Does there exist a closed set which is an intersection of a collection of infinite open sets?

analysis

share|cite|improve this question

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

332

New contributor

Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as off-topic by Eevee Trainer, Shalop, Shailesh, user21820, José Carlos Santos Mar 18 at 10:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Shalop, Shailesh, user21820, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  Intersect $(-tfrac1n, tfrac1n)$ for $n = 1, 2, ldots$ and consider what set you get
  $endgroup$
  – Brevan Ellefsen
  Mar 18 at 5:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Oh it will get $0$
  $endgroup$
  – Tony Tong
  Mar 18 at 5:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrevanEllefsen: The question only asks for a "collection of infinite open sets", so just the singleton $(0,1)$ suffices. But Tony, please read how to ask a good question.
  $endgroup$
  – user21820
  Mar 18 at 9:23
  
  

  
  
  
  

add a comment | 

$begingroup$

Does there exist a closed set which is an intersection of a collection of infinite open sets?

analysis

share|cite|improve this question

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

332

New contributor

Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

332

New contributor

Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

332

New contributor

Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

edited Mar 18 at 8:39

YuiTo Cheng

2,0592837

asked Mar 18 at 5:50

Tony TongTony Tong

332

New contributor

Tony Tong is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 18 at 5:50

Tony TongTony Tong

332

2 Answers
2

active

oldest

votes

8

$begingroup$

$$mathbbRcapmathbbRcapmathbbRcapcdots$$

share|cite|improve this answer

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But R is an open set, the intersection is also R so it is still an open set
  $endgroup$
  – Tony Tong
  Mar 18 at 5:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And also closed
  $endgroup$
  – Keen-ameteur
  Mar 18 at 5:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  While extremely simple, this example has the unfortunate side effect of also being open, which could further confound the OP who seems to be wondering why the intersection of open sets is not open in general (admittedly, the OP is probably also drawing a false dichotomy between open and closed sets, so I suppose this helps with that)
  $endgroup$
  – Brevan Ellefsen
  Mar 18 at 5:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrevanEllefsen: +1, but I couldn't resist... :-)
  $endgroup$
  – parsiad
  Mar 18 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That's a bad answer from the pedagogical point of view… but it's really funny too!
  $endgroup$
  – José Carlos Santos
  Mar 18 at 6:56
  

  
  
  
  

add a comment | 

3

$begingroup$

As other answers and comments point out, this does indeed hold. In fact, every closed set in $mathbb R$ is a (countable) intersection of open sets. This holds in any metric space $(X,d)$. Here is a proof in the general metric case:

Suppose $Asubset X$ is closed. For each $n$, let
$$
U_n :=bigcup_ain A B(a,1/n).
$$
Note that $U_n$ is open (union of open sets). We claim that
$$
A = bigcap_ninmathbb N U_n.
$$
First note that $Asubset bigcap U_n$ since $Asubset U_n$ for each $n$ (by construction!). Also, if $xinbigcap U_n$, then for each $n$ we have some $a_n$ so that $xin B(a_n, 1/n)$, i.e. $d(x,a_n) < 1/n$ for each $n$. This implies that $a_nrightarrow x$ as $nrightarrowinfty$. Since $A$ is closed, it follows that $xin A$. QED.

share|cite|improve this answer

edited Mar 18 at 7:04

answered Mar 18 at 6:59

o.h.o.h.

6817

$endgroup$

add a comment | 

8

$begingroup$

$$mathbbRcapmathbbRcapmathbbRcapcdots$$

share|cite|improve this answer

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But R is an open set, the intersection is also R so it is still an open set
  $endgroup$
  – Tony Tong
  Mar 18 at 5:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And also closed
  $endgroup$
  – Keen-ameteur
  Mar 18 at 5:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  While extremely simple, this example has the unfortunate side effect of also being open, which could further confound the OP who seems to be wondering why the intersection of open sets is not open in general (admittedly, the OP is probably also drawing a false dichotomy between open and closed sets, so I suppose this helps with that)
  $endgroup$
  – Brevan Ellefsen
  Mar 18 at 5:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrevanEllefsen: +1, but I couldn't resist... :-)
  $endgroup$
  – parsiad
  Mar 18 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That's a bad answer from the pedagogical point of view… but it's really funny too!
  $endgroup$
  – José Carlos Santos
  Mar 18 at 6:56
  

  
  
  
  

add a comment | 

8

$begingroup$

$$mathbbRcapmathbbRcapmathbbRcapcdots$$

share|cite|improve this answer

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But R is an open set, the intersection is also R so it is still an open set
  $endgroup$
  – Tony Tong
  Mar 18 at 5:54
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  And also closed
  $endgroup$
  – Keen-ameteur
  Mar 18 at 5:55
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  While extremely simple, this example has the unfortunate side effect of also being open, which could further confound the OP who seems to be wondering why the intersection of open sets is not open in general (admittedly, the OP is probably also drawing a false dichotomy between open and closed sets, so I suppose this helps with that)
  $endgroup$
  – Brevan Ellefsen
  Mar 18 at 5:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BrevanEllefsen: +1, but I couldn't resist... :-)
  $endgroup$
  – parsiad
  Mar 18 at 6:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That's a bad answer from the pedagogical point of view… but it's really funny too!
  $endgroup$
  – José Carlos Santos
  Mar 18 at 6:56
  

  
  
  
  

add a comment | 

$begingroup$

$$mathbbRcapmathbbRcapmathbbRcapcdots$$

share|cite|improve this answer

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

$endgroup$

share|cite|improve this answer

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

answered Mar 18 at 5:53

parsiadparsiad

18.5k32453

3

$begingroup$

As other answers and comments point out, this does indeed hold. In fact, every closed set in $mathbb R$ is a (countable) intersection of open sets. This holds in any metric space $(X,d)$. Here is a proof in the general metric case:

Suppose $Asubset X$ is closed. For each $n$, let
$$
U_n :=bigcup_ain A B(a,1/n).
$$
Note that $U_n$ is open (union of open sets). We claim that
$$
A = bigcap_ninmathbb N U_n.
$$
First note that $Asubset bigcap U_n$ since $Asubset U_n$ for each $n$ (by construction!). Also, if $xinbigcap U_n$, then for each $n$ we have some $a_n$ so that $xin B(a_n, 1/n)$, i.e. $d(x,a_n) < 1/n$ for each $n$. This implies that $a_nrightarrow x$ as $nrightarrowinfty$. Since $A$ is closed, it follows that $xin A$. QED.

share|cite|improve this answer

edited Mar 18 at 7:04

answered Mar 18 at 6:59

o.h.o.h.

6817

$endgroup$

add a comment | 

3

$begingroup$

As other answers and comments point out, this does indeed hold. In fact, every closed set in $mathbb R$ is a (countable) intersection of open sets. This holds in any metric space $(X,d)$. Here is a proof in the general metric case:

Suppose $Asubset X$ is closed. For each $n$, let
$$
U_n :=bigcup_ain A B(a,1/n).
$$
Note that $U_n$ is open (union of open sets). We claim that
$$
A = bigcap_ninmathbb N U_n.
$$
First note that $Asubset bigcap U_n$ since $Asubset U_n$ for each $n$ (by construction!). Also, if $xinbigcap U_n$, then for each $n$ we have some $a_n$ so that $xin B(a_n, 1/n)$, i.e. $d(x,a_n) < 1/n$ for each $n$. This implies that $a_nrightarrow x$ as $nrightarrowinfty$. Since $A$ is closed, it follows that $xin A$. QED.

share|cite|improve this answer

edited Mar 18 at 7:04

answered Mar 18 at 6:59

o.h.o.h.

6817

$endgroup$

add a comment | 

$begingroup$

As other answers and comments point out, this does indeed hold. In fact, every closed set in $mathbb R$ is a (countable) intersection of open sets. This holds in any metric space $(X,d)$. Here is a proof in the general metric case:

Suppose $Asubset X$ is closed. For each $n$, let
$$
U_n :=bigcup_ain A B(a,1/n).
$$
Note that $U_n$ is open (union of open sets). We claim that
$$
A = bigcap_ninmathbb N U_n.
$$
First note that $Asubset bigcap U_n$ since $Asubset U_n$ for each $n$ (by construction!). Also, if $xinbigcap U_n$, then for each $n$ we have some $a_n$ so that $xin B(a_n, 1/n)$, i.e. $d(x,a_n) < 1/n$ for each $n$. This implies that $a_nrightarrow x$ as $nrightarrowinfty$. Since $A$ is closed, it follows that $xin A$. QED.

share|cite|improve this answer

edited Mar 18 at 7:04

answered Mar 18 at 6:59

o.h.o.h.

6817

$endgroup$

share|cite|improve this answer

edited Mar 18 at 7:04

answered Mar 18 at 6:59

o.h.o.h.

6817

answered Mar 18 at 6:59

o.h.o.h.

6817

answered Mar 18 at 6:59

o.h.o.h.

6817