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How to define the density on a subspace of a topological one

0

$begingroup$

Assume $X$ is a topological space and $Asubseteq X$.
Then $A$ is dense in $X$ iff $overlineA=X$.
Now, suppose $Asubseteq Bsubseteq X$.

How can I define the density of $A$ in $B$?

As I understand it, a reasonable way to overcome this difficulty is to require $overlineAcap B=B$ or $overlineAcap B=B$, where we are using the closure operator of $X$.

I guess the second one is more adeguate, but it is only a feeling.

general-topology

share|cite|improve this question

edited Mar 27 at 16:29

LBJFS

asked Mar 27 at 16:25

LBJFSLBJFS

391112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $Acap B = A$.
  $endgroup$
  – amsmath
  Mar 27 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I feel so stupid... at this point is the first one the actual definition?
  $endgroup$
  – LBJFS
  Mar 27 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The second is at least not the one because it would imply that $B$ is closed.
  $endgroup$
  – amsmath
  Mar 27 at 16:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are absolutely right once again
  $endgroup$
  – LBJFS
  Mar 27 at 16:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
  $endgroup$
  – amsmath
  Mar 27 at 16:34
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

Assume $X$ is a topological space and $Asubseteq X$.
Then $A$ is dense in $X$ iff $overlineA=X$.
Now, suppose $Asubseteq Bsubseteq X$.

How can I define the density of $A$ in $B$?

As I understand it, a reasonable way to overcome this difficulty is to require $overlineAcap B=B$ or $overlineAcap B=B$, where we are using the closure operator of $X$.

I guess the second one is more adeguate, but it is only a feeling.

general-topology

share|cite|improve this question

edited Mar 27 at 16:29

LBJFS

asked Mar 27 at 16:25

LBJFSLBJFS

391112

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $Acap B = A$.
  $endgroup$
  – amsmath
  Mar 27 at 16:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, I feel so stupid... at this point is the first one the actual definition?
  $endgroup$
  – LBJFS
  Mar 27 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The second is at least not the one because it would imply that $B$ is closed.
  $endgroup$
  – amsmath
  Mar 27 at 16:31
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are absolutely right once again
  $endgroup$
  – LBJFS
  Mar 27 at 16:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
  $endgroup$
  – amsmath
  Mar 27 at 16:34
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Assume $X$ is a topological space and $Asubseteq X$.
Then $A$ is dense in $X$ iff $overlineA=X$.
Now, suppose $Asubseteq Bsubseteq X$.

How can I define the density of $A$ in $B$?

As I understand it, a reasonable way to overcome this difficulty is to require $overlineAcap B=B$ or $overlineAcap B=B$, where we are using the closure operator of $X$.

I guess the second one is more adeguate, but it is only a feeling.

general-topology

share|cite|improve this question

edited Mar 27 at 16:29

LBJFS

asked Mar 27 at 16:25

LBJFSLBJFS

391112

$endgroup$

share|cite|improve this question

edited Mar 27 at 16:29

LBJFS

asked Mar 27 at 16:25

LBJFSLBJFS

391112

share|cite|improve this question

edited Mar 27 at 16:29

LBJFS

asked Mar 27 at 16:25

LBJFSLBJFS

391112

asked Mar 27 at 16:25

LBJFSLBJFS

391112

asked Mar 27 at 16:25

LBJFSLBJFS

391112

2 Answers
2

active

oldest

votes

1

$begingroup$

By definition, $A$ is dense in $B$ if for any $bin B$ and any open $U$ with $bin U$ you can find $ain A$ such that $ain Ucap B$, i.e., $ain Acap U$.

Assume that $overline Acap B = B$ and let $bin B$, $U$ open with $bin U$. Then, since $bin overline A$, there exists $ain Acap U$.

Assume that $A$ is dense in $B$ and let $bin B$. Then for any open $U$ with $bin U$ we find $ain Acap U$. This implies $bin overline A$. Hence, we have $overline Acap B = B$.

share|cite|improve this answer

answered Mar 27 at 16:48

amsmathamsmath

3,387421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
  $endgroup$
  – LBJFS
  Mar 27 at 16:54
  

  
  
  
  

add a comment | 

1

$begingroup$

$A$ is dense in $B$ iff $operatornamecl_B(A)=B$ so iff $operatornamecl_X(A) cap B = B$ (by the formula that relates closure in subspaces to that in the large space) iff $B subseteq operatornamecl_X(A)$. So it suffices to check the closure of $A$ in the whole space and check if it contains $B$.

It's intuitive too: $A$ is dense in $B$ iff we can "reach" all points of $B$ by points of $A$ (formalisable by nets, e.g.) and this also can be restated as $B subseteq operatornamecl_X(A)$

share|cite|improve this answer

edited Mar 27 at 18:55

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, now I see. This is the natural way I was looking for. Thank you very much!
  $endgroup$
  – LBJFS
  Mar 27 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He just restated your statement that $A$ is dense in $B$ iff $overline Acap B = B$. And "this last condition also restates that same fact." I proved in my answer.
  $endgroup$
  – amsmath
  Mar 27 at 18:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean, but he stated explicitly why the formula holds: it was sufficient to recall the definition of subspace closure. Maybe it is trivial, but this is why everything works.
  $endgroup$
  – LBJFS
  Mar 27 at 18:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You mean "relative closure". But this has to be proved using the relative topology. That's what you have first. And I proved in my answer that the relative closure of $A$ is $overline Acap B$. If you knew that already, your question seems odd.
  $endgroup$
  – amsmath
  Mar 27 at 18:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @LBJFS I did some time ago; a minor edit.
  $endgroup$
  – Henno Brandsma
  Mar 27 at 22:23
  

  
  
  
  

 | 
show 4 more comments

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1

$begingroup$

By definition, $A$ is dense in $B$ if for any $bin B$ and any open $U$ with $bin U$ you can find $ain A$ such that $ain Ucap B$, i.e., $ain Acap U$.

Assume that $overline Acap B = B$ and let $bin B$, $U$ open with $bin U$. Then, since $bin overline A$, there exists $ain Acap U$.

Assume that $A$ is dense in $B$ and let $bin B$. Then for any open $U$ with $bin U$ we find $ain Acap U$. This implies $bin overline A$. Hence, we have $overline Acap B = B$.

share|cite|improve this answer

answered Mar 27 at 16:48

amsmathamsmath

3,387421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
  $endgroup$
  – LBJFS
  Mar 27 at 16:54
  

  
  
  
  

add a comment | 

1

$begingroup$

By definition, $A$ is dense in $B$ if for any $bin B$ and any open $U$ with $bin U$ you can find $ain A$ such that $ain Ucap B$, i.e., $ain Acap U$.

Assume that $overline Acap B = B$ and let $bin B$, $U$ open with $bin U$. Then, since $bin overline A$, there exists $ain Acap U$.

Assume that $A$ is dense in $B$ and let $bin B$. Then for any open $U$ with $bin U$ we find $ain Acap U$. This implies $bin overline A$. Hence, we have $overline Acap B = B$.

share|cite|improve this answer

answered Mar 27 at 16:48

amsmathamsmath

3,387421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
  $endgroup$
  – LBJFS
  Mar 27 at 16:54
  

  
  
  
  

add a comment | 

$begingroup$

By definition, $A$ is dense in $B$ if for any $bin B$ and any open $U$ with $bin U$ you can find $ain A$ such that $ain Ucap B$, i.e., $ain Acap U$.

Assume that $overline Acap B = B$ and let $bin B$, $U$ open with $bin U$. Then, since $bin overline A$, there exists $ain Acap U$.

Assume that $A$ is dense in $B$ and let $bin B$. Then for any open $U$ with $bin U$ we find $ain Acap U$. This implies $bin overline A$. Hence, we have $overline Acap B = B$.

share|cite|improve this answer

answered Mar 27 at 16:48

amsmathamsmath

3,387421

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:48

amsmathamsmath

3,387421

answered Mar 27 at 16:48

amsmathamsmath

3,387421

answered Mar 27 at 16:48

amsmathamsmath

3,387421

1

$begingroup$

$A$ is dense in $B$ iff $operatornamecl_B(A)=B$ so iff $operatornamecl_X(A) cap B = B$ (by the formula that relates closure in subspaces to that in the large space) iff $B subseteq operatornamecl_X(A)$. So it suffices to check the closure of $A$ in the whole space and check if it contains $B$.

It's intuitive too: $A$ is dense in $B$ iff we can "reach" all points of $B$ by points of $A$ (formalisable by nets, e.g.) and this also can be restated as $B subseteq operatornamecl_X(A)$

share|cite|improve this answer

edited Mar 27 at 18:55

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, now I see. This is the natural way I was looking for. Thank you very much!
  $endgroup$
  – LBJFS
  Mar 27 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He just restated your statement that $A$ is dense in $B$ iff $overline Acap B = B$. And "this last condition also restates that same fact." I proved in my answer.
  $endgroup$
  – amsmath
  Mar 27 at 18:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean, but he stated explicitly why the formula holds: it was sufficient to recall the definition of subspace closure. Maybe it is trivial, but this is why everything works.
  $endgroup$
  – LBJFS
  Mar 27 at 18:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You mean "relative closure". But this has to be proved using the relative topology. That's what you have first. And I proved in my answer that the relative closure of $A$ is $overline Acap B$. If you knew that already, your question seems odd.
  $endgroup$
  – amsmath
  Mar 27 at 18:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @LBJFS I did some time ago; a minor edit.
  $endgroup$
  – Henno Brandsma
  Mar 27 at 22:23
  

  
  
  
  

 | 
show 4 more comments

1

$begingroup$

$A$ is dense in $B$ iff $operatornamecl_B(A)=B$ so iff $operatornamecl_X(A) cap B = B$ (by the formula that relates closure in subspaces to that in the large space) iff $B subseteq operatornamecl_X(A)$. So it suffices to check the closure of $A$ in the whole space and check if it contains $B$.

It's intuitive too: $A$ is dense in $B$ iff we can "reach" all points of $B$ by points of $A$ (formalisable by nets, e.g.) and this also can be restated as $B subseteq operatornamecl_X(A)$

share|cite|improve this answer

edited Mar 27 at 18:55

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah, now I see. This is the natural way I was looking for. Thank you very much!
  $endgroup$
  – LBJFS
  Mar 27 at 17:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  He just restated your statement that $A$ is dense in $B$ iff $overline Acap B = B$. And "this last condition also restates that same fact." I proved in my answer.
  $endgroup$
  – amsmath
  Mar 27 at 18:15
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I see what you mean, but he stated explicitly why the formula holds: it was sufficient to recall the definition of subspace closure. Maybe it is trivial, but this is why everything works.
  $endgroup$
  – LBJFS
  Mar 27 at 18:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You mean "relative closure". But this has to be proved using the relative topology. That's what you have first. And I proved in my answer that the relative closure of $A$ is $overline Acap B$. If you knew that already, your question seems odd.
  $endgroup$
  – amsmath
  Mar 27 at 18:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @LBJFS I did some time ago; a minor edit.
  $endgroup$
  – Henno Brandsma
  Mar 27 at 22:23
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

$A$ is dense in $B$ iff $operatornamecl_B(A)=B$ so iff $operatornamecl_X(A) cap B = B$ (by the formula that relates closure in subspaces to that in the large space) iff $B subseteq operatornamecl_X(A)$. So it suffices to check the closure of $A$ in the whole space and check if it contains $B$.

It's intuitive too: $A$ is dense in $B$ iff we can "reach" all points of $B$ by points of $A$ (formalisable by nets, e.g.) and this also can be restated as $B subseteq operatornamecl_X(A)$

share|cite|improve this answer

edited Mar 27 at 18:55

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

$endgroup$

share|cite|improve this answer

edited Mar 27 at 18:55

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127

answered Mar 27 at 17:40

Henno BrandsmaHenno Brandsma

117k349127