How to define the density on a subspace of a topological one The Next CEO of Stack OverflowThe topology produced by indexing subsets of a topological space.Density, Irreducible Topological SpaceAlgebraic (Kuratowski axiomatic) proof of a simple topological statementCondition ici=ic on a topological space is equivalent to if each dense set has dense interior in the space.Second category subset vs subspaceTwo definitions of one set being dense in the other$A$ dense in $X$, $O$ open in $X$Tarski's notion of $E$-spaceBasis of topological space and topological subspaceEquivalent definition of irreducible topological subspace.
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How to define the density on a subspace of a topological one
The Next CEO of Stack OverflowThe topology produced by indexing subsets of a topological space.Density, Irreducible Topological SpaceAlgebraic (Kuratowski axiomatic) proof of a simple topological statementCondition ici=ic on a topological space is equivalent to if each dense set has dense interior in the space.Second category subset vs subspaceTwo definitions of one set being dense in the other$A$ dense in $X$, $O$ open in $X$Tarski's notion of $E$-spaceBasis of topological space and topological subspaceEquivalent definition of irreducible topological subspace.
$begingroup$
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
asked 2 days ago
LBJFSLBJFS
31311
$endgroup$
|
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
asked 2 days ago
LBJFSLBJFS
31311
$endgroup$
1
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago
|
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
asked 2 days ago
LBJFSLBJFS
31311
$endgroup$
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
general-topology
asked 2 days ago
LBJFSLBJFS
31311
asked 2 days ago
LBJFSLBJFS
31311
edited 2 days ago
LBJFS
asked 2 days ago
LBJFSLBJFS
31311
asked 2 days ago
LBJFSLBJFS
31311
asked 2 days ago
LBJFSLBJFS
31311
31311
1
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago
|
show 2 more comments
1
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago
1
1
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
1
1
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
1
1
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 $overline Acap B = B$.
answered 2 days ago
amsmathamsmath
3,278419
$endgroup$
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
add a comment |
$begingroup$
$A$ 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 $B subseteq operatornamecl_X(A)$
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
|
show 4 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $overline Acap B = B$.
answered 2 days ago
amsmathamsmath
3,278419
$endgroup$
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
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 $overline Acap B = B$.
answered 2 days ago
amsmathamsmath
3,278419
$endgroup$
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
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 $overline Acap B = B$.
answered 2 days ago
amsmathamsmath
3,278419
$endgroup$
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 $overline Acap B = B$.
answered 2 days ago
amsmathamsmath
3,278419
answered 2 days ago
amsmathamsmath
3,278419
answered 2 days ago
amsmathamsmath
3,278419
answered 2 days ago
amsmathamsmath
3,278419
3,278419
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
add a comment |
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
$begingroup$
Ok, you checked the agreement of this definition with the equivalent notions in $X$. Thank you very much.
$endgroup$
– LBJFS
2 days ago
add a comment |
$begingroup$
$A$ 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 $B subseteq operatornamecl_X(A)$
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
|
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 $B subseteq operatornamecl_X(A)$
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
|
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 $B subseteq operatornamecl_X(A)$
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
$endgroup$
$A$ 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 $B subseteq operatornamecl_X(A)$
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
edited 2 days ago
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
answered 2 days ago
Henno BrandsmaHenno Brandsma
114k348124
114k348124
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
|
show 4 more comments
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
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
2 days ago
$begingroup$
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
2 days ago
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
2 days ago
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
$begingroup$
Ah, now I see. This is the natural way I was looking for. Thank you very much!
$endgroup$
– LBJFS
2 days ago
1
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
$begingroup$
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
2 days ago
1
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
2 days ago
$begingroup$
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
2 days ago
1
1
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
$begingroup$
@LBJFS I did some time ago; a minor edit.
$endgroup$
– Henno Brandsma
2 days ago
|
show 4 more comments
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1
$begingroup$
Note that $Acap B = A$.
$endgroup$
– amsmath
2 days ago
$begingroup$
Ah, I feel so stupid... at this point is the first one the actual definition?
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
The second is at least not the one because it would imply that $B$ is closed.
$endgroup$
– amsmath
2 days ago
$begingroup$
You are absolutely right once again
$endgroup$
– LBJFS
2 days ago
1
$begingroup$
It is correct that $A$ is dense in $B$ if and only if $overline Acap B = B$.
$endgroup$
– amsmath
2 days ago