Fdtxjr

Multi tool use

What is the topology associated with the algebras for the ultrafilter monad?

Amount of permutations on an NxNxN Rubik's Cube

How to install press fit bottom bracket into new frame

How to compare two different files line by line in unix?

Is it possible for SQL statements to execute concurrently within a single session in SQL Server?

Is there hard evidence that the grant peer review system performs significantly better than random?

Why should I vote and accept answers?

Most bit efficient text communication method?

Question about debouncing - delay of state change

Did Deadpool rescue all of the X-Force?

How can I reduce the gap between left and right of cdot with a macro?

What would you call this weird metallic apparatus that allows you to lift people?

If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?

What is "gratricide"?

What is a fractional matching?

As a beginner, should I get a Squier Strat with a SSS config or a HSS?

How does light 'choose' between wave and particle behaviour?

How to write the following sign?

Morning, Afternoon, Night Kanji

Denied boarding although I have proper visa and documentation. To whom should I make a complaint?

Do I really need to have a message in a novel to appeal to readers?

How come Sam didn't become Lord of Horn Hill?

Can the Great Weapon Master feat's damage bonus and accuracy penalty apply to attacks from the Spiritual Weapon spell?

Why wasn't DOSKEY integrated with COMMAND.COM?

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

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164714%2fhow-to-define-the-density-on-a-subspace-of-a-topological-one%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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