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Intersection of nested closed sets in Hausdorf topological space
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I am trying to prove the following statement, but starting to doubt its correctness.
Suppose that $H$ is a Hausdorf topological space (I am formulating generally, though my specific case is $H=S'(R)$ - a space of tempered distributions).
Suppose I have a set of nested subsets $Omega_i subseteq H$ and $Omega_i supseteq Omega_i+1$ and by $overlineOmega$ we denote a sequential closure of $Omegasubseteq H$. Is it true that:
$xin cap_i=1^infty overlineOmega_i$ if and only if there is sequence $x_i$ such that $x_i rightarrow x, x_iin Omega_i$.
The fact that from $x_i rightarrow x, x_iin Omega_i$ we can deduce $xin cap_i=1^infty overlineOmega_i$ is obvious. The opposite is problematic.
I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $xin cap_i=1^infty overlineOmega_i$ is fixed. Then I define
$R_i = exists N, forall n>N: x_n in Omega_i, x_n rightarrow x$ (a set of sequences that tend to $x$ and is in $Omega_i$ starting from some index). It is easy to see that $R_i$ is also nested: $R_isupseteq R_i+1$.
Then the wanted statement is equivalent to $cap_i R_i ne emptyset$.
The problem now is that I need compactness of $R_i$ in order to apply Kantor's theorem, but I stuck at this step.
real-analysis general-topology functional-analysis
asked Aug 3 at 14:07
redkin77
113
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2
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I am trying to prove the following statement, but starting to doubt its correctness.
Suppose that $H$ is a Hausdorf topological space (I am formulating generally, though my specific case is $H=S'(R)$ - a space of tempered distributions).
Suppose I have a set of nested subsets $Omega_i subseteq H$ and $Omega_i supseteq Omega_i+1$ and by $overlineOmega$ we denote a sequential closure of $Omegasubseteq H$. Is it true that:
$xin cap_i=1^infty overlineOmega_i$ if and only if there is sequence $x_i$ such that $x_i rightarrow x, x_iin Omega_i$.
The fact that from $x_i rightarrow x, x_iin Omega_i$ we can deduce $xin cap_i=1^infty overlineOmega_i$ is obvious. The opposite is problematic.
I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $xin cap_i=1^infty overlineOmega_i$ is fixed. Then I define
$R_i = exists N, forall n>N: x_n in Omega_i, x_n rightarrow x$ (a set of sequences that tend to $x$ and is in $Omega_i$ starting from some index). It is easy to see that $R_i$ is also nested: $R_isupseteq R_i+1$.
Then the wanted statement is equivalent to $cap_i R_i ne emptyset$.
The problem now is that I need compactness of $R_i$ in order to apply Kantor's theorem, but I stuck at this step.
real-analysis general-topology functional-analysis
asked Aug 3 at 14:07
redkin77
113
You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to prove the following statement, but starting to doubt its correctness.
Suppose that $H$ is a Hausdorf topological space (I am formulating generally, though my specific case is $H=S'(R)$ - a space of tempered distributions).
Suppose I have a set of nested subsets $Omega_i subseteq H$ and $Omega_i supseteq Omega_i+1$ and by $overlineOmega$ we denote a sequential closure of $Omegasubseteq H$. Is it true that:
$xin cap_i=1^infty overlineOmega_i$ if and only if there is sequence $x_i$ such that $x_i rightarrow x, x_iin Omega_i$.
The fact that from $x_i rightarrow x, x_iin Omega_i$ we can deduce $xin cap_i=1^infty overlineOmega_i$ is obvious. The opposite is problematic.
I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $xin cap_i=1^infty overlineOmega_i$ is fixed. Then I define
$R_i = exists N, forall n>N: x_n in Omega_i, x_n rightarrow x$ (a set of sequences that tend to $x$ and is in $Omega_i$ starting from some index). It is easy to see that $R_i$ is also nested: $R_isupseteq R_i+1$.
Then the wanted statement is equivalent to $cap_i R_i ne emptyset$.
The problem now is that I need compactness of $R_i$ in order to apply Kantor's theorem, but I stuck at this step.
real-analysis general-topology functional-analysis
asked Aug 3 at 14:07
redkin77
113
I am trying to prove the following statement, but starting to doubt its correctness.
Suppose that $H$ is a Hausdorf topological space (I am formulating generally, though my specific case is $H=S'(R)$ - a space of tempered distributions).
Suppose I have a set of nested subsets $Omega_i subseteq H$ and $Omega_i supseteq Omega_i+1$ and by $overlineOmega$ we denote a sequential closure of $Omegasubseteq H$. Is it true that:
$xin cap_i=1^infty overlineOmega_i$ if and only if there is sequence $x_i$ such that $x_i rightarrow x, x_iin Omega_i$.
The fact that from $x_i rightarrow x, x_iin Omega_i$ we can deduce $xin cap_i=1^infty overlineOmega_i$ is obvious. The opposite is problematic.
I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $xin cap_i=1^infty overlineOmega_i$ is fixed. Then I define
$R_i = exists N, forall n>N: x_n in Omega_i, x_n rightarrow x$ (a set of sequences that tend to $x$ and is in $Omega_i$ starting from some index). It is easy to see that $R_i$ is also nested: $R_isupseteq R_i+1$.
Then the wanted statement is equivalent to $cap_i R_i ne emptyset$.
The problem now is that I need compactness of $R_i$ in order to apply Kantor's theorem, but I stuck at this step.
real-analysis general-topology functional-analysis
asked Aug 3 at 14:07
redkin77
113
edited Aug 3 at 15:27
asked Aug 3 at 14:07
redkin77
113
asked Aug 3 at 14:07
redkin77
113
asked Aug 3 at 14:07
redkin77
113
113
You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55
 |Â
show 1 more comment
You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55
You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55
 |Â
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For a counterexample using sequential closures, consider the topological space whose underlying set is $mathbbN^2 sqcup x_0 $, and with the topology such that $U$ is open if and only if $x_0 notin U$ or for some function $f : mathbbN to mathbbN$, $ (x, y) in mathbbN^2 mid y > f(x) subseteq U$.
Now, let $Omega_n := (x, y) in mathbbN^2 mid x ge n $. Then $x_0$ is in the sequential closure of $Omega_n$ for each $n$ since $(n, m) to x_0$ as $m to infty$. On the other hand, if we have any sequence $(x_n, y_n) in Omega_n$, then $x_n to infty$ as $n to infty$. Using this, it is possible to construct a function $f : mathbbN to mathbbN$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) notto x_0$ as $n to infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
add a comment |Â
up vote
1
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The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true.
If $x in cap_i barOmega_i$, then for each $i$, there is a sequence $(y_i,n)_n$ in $Omega_i$ that converges to $x$, since $x in barOmega_i$. Then what about the following sequence :
$$
x_j = y_j,j ; forall j ?
$$
Well for each $j$, $x_j = y_j,j$ is in $Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_0,0
$$
and
$$
x_j+1 = y_j+1,i_j
$$
where $i_j$ is the smallest integer such that $d(y_j+1,i_j,x)<2^-j$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_k,l)$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential :
Suppose the statement is true, consider a familly $Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
answered Aug 3 at 16:13
jeanmfischer
478212
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
 |Â
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2 Answers
2
active
oldest
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active
oldest
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active
oldest
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up vote
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For a counterexample using sequential closures, consider the topological space whose underlying set is $mathbbN^2 sqcup x_0 $, and with the topology such that $U$ is open if and only if $x_0 notin U$ or for some function $f : mathbbN to mathbbN$, $ (x, y) in mathbbN^2 mid y > f(x) subseteq U$.
Now, let $Omega_n := (x, y) in mathbbN^2 mid x ge n $. Then $x_0$ is in the sequential closure of $Omega_n$ for each $n$ since $(n, m) to x_0$ as $m to infty$. On the other hand, if we have any sequence $(x_n, y_n) in Omega_n$, then $x_n to infty$ as $n to infty$. Using this, it is possible to construct a function $f : mathbbN to mathbbN$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) notto x_0$ as $n to infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
add a comment |Â
up vote
2
down vote
For a counterexample using sequential closures, consider the topological space whose underlying set is $mathbbN^2 sqcup x_0 $, and with the topology such that $U$ is open if and only if $x_0 notin U$ or for some function $f : mathbbN to mathbbN$, $ (x, y) in mathbbN^2 mid y > f(x) subseteq U$.
Now, let $Omega_n := (x, y) in mathbbN^2 mid x ge n $. Then $x_0$ is in the sequential closure of $Omega_n$ for each $n$ since $(n, m) to x_0$ as $m to infty$. On the other hand, if we have any sequence $(x_n, y_n) in Omega_n$, then $x_n to infty$ as $n to infty$. Using this, it is possible to construct a function $f : mathbbN to mathbbN$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) notto x_0$ as $n to infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For a counterexample using sequential closures, consider the topological space whose underlying set is $mathbbN^2 sqcup x_0 $, and with the topology such that $U$ is open if and only if $x_0 notin U$ or for some function $f : mathbbN to mathbbN$, $ (x, y) in mathbbN^2 mid y > f(x) subseteq U$.
Now, let $Omega_n := (x, y) in mathbbN^2 mid x ge n $. Then $x_0$ is in the sequential closure of $Omega_n$ for each $n$ since $(n, m) to x_0$ as $m to infty$. On the other hand, if we have any sequence $(x_n, y_n) in Omega_n$, then $x_n to infty$ as $n to infty$. Using this, it is possible to construct a function $f : mathbbN to mathbbN$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) notto x_0$ as $n to infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
For a counterexample using sequential closures, consider the topological space whose underlying set is $mathbbN^2 sqcup x_0 $, and with the topology such that $U$ is open if and only if $x_0 notin U$ or for some function $f : mathbbN to mathbbN$, $ (x, y) in mathbbN^2 mid y > f(x) subseteq U$.
Now, let $Omega_n := (x, y) in mathbbN^2 mid x ge n $. Then $x_0$ is in the sequential closure of $Omega_n$ for each $n$ since $(n, m) to x_0$ as $m to infty$. On the other hand, if we have any sequence $(x_n, y_n) in Omega_n$, then $x_n to infty$ as $n to infty$. Using this, it is possible to construct a function $f : mathbbN to mathbbN$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) notto x_0$ as $n to infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
answered Aug 3 at 17:45
Daniel Schepler
6,6281513
6,6281513
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
add a comment |Â
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
Nice counterexample, I am so jealous of it :)
– jeanmfischer
Aug 3 at 17:59
1
1
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general...
– Daniel Schepler
Aug 3 at 18:00
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Indeed, very beautiful! I guess I need to add some natural constraints on $Omega_i$ in order to prove the statement. Thank you a lot.
– redkin77
Aug 3 at 18:04
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
Is this the space where non idempotency is realised ?
– jeanmfischer
Aug 3 at 18:46
1
1
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
@jeanmfischer No, the space for that is $ x_mn sqcup y_m sqcup z $ where basic neighborhoods of $x_mn$ are $ x_mn $, basic neighborhoods of $y_m$ are $ x_mn mid n > n_0 cup y_m $, basic neighborhoods of $z$ are $ x_mn mid n > f(m) wedge m > m_0 cup y_m mid m > m_0 cup z $ - so $x_mn to y_m$ as $nto infty$ and $y_m to z$ as $mto infty$ but no sequence within $ x_mn $ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$.
– Daniel Schepler
Aug 3 at 18:57
add a comment |Â
up vote
1
down vote
The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true.
If $x in cap_i barOmega_i$, then for each $i$, there is a sequence $(y_i,n)_n$ in $Omega_i$ that converges to $x$, since $x in barOmega_i$. Then what about the following sequence :
$$
x_j = y_j,j ; forall j ?
$$
Well for each $j$, $x_j = y_j,j$ is in $Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_0,0
$$
and
$$
x_j+1 = y_j+1,i_j
$$
where $i_j$ is the smallest integer such that $d(y_j+1,i_j,x)<2^-j$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_k,l)$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential :
Suppose the statement is true, consider a familly $Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
answered Aug 3 at 16:13
jeanmfischer
478212
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
 |Â
show 8 more comments
up vote
1
down vote
The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true.
If $x in cap_i barOmega_i$, then for each $i$, there is a sequence $(y_i,n)_n$ in $Omega_i$ that converges to $x$, since $x in barOmega_i$. Then what about the following sequence :
$$
x_j = y_j,j ; forall j ?
$$
Well for each $j$, $x_j = y_j,j$ is in $Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_0,0
$$
and
$$
x_j+1 = y_j+1,i_j
$$
where $i_j$ is the smallest integer such that $d(y_j+1,i_j,x)<2^-j$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_k,l)$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential :
Suppose the statement is true, consider a familly $Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
answered Aug 3 at 16:13
jeanmfischer
478212
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
 |Â
show 8 more comments
up vote
1
down vote
up vote
1
down vote
The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true.
If $x in cap_i barOmega_i$, then for each $i$, there is a sequence $(y_i,n)_n$ in $Omega_i$ that converges to $x$, since $x in barOmega_i$. Then what about the following sequence :
$$
x_j = y_j,j ; forall j ?
$$
Well for each $j$, $x_j = y_j,j$ is in $Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_0,0
$$
and
$$
x_j+1 = y_j+1,i_j
$$
where $i_j$ is the smallest integer such that $d(y_j+1,i_j,x)<2^-j$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_k,l)$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential :
Suppose the statement is true, consider a familly $Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
answered Aug 3 at 16:13
jeanmfischer
478212
The space of distributions is not even a sequential space (indeed not first countable) so what you are trying to prove has little hope of beeing true.
If $x in cap_i barOmega_i$, then for each $i$, there is a sequence $(y_i,n)_n$ in $Omega_i$ that converges to $x$, since $x in barOmega_i$. Then what about the following sequence :
$$
x_j = y_j,j ; forall j ?
$$
Well for each $j$, $x_j = y_j,j$ is in $Omega_j$ by construction.
What about convergence ? Here maybe you need more than just Hausdorff and sequential, lets use metric !
Instead of using the diagonal sequence, build the sequence by induction :
$$
x_0 = y_0,0
$$
and
$$
x_j+1 = y_j+1,i_j
$$
where $i_j$ is the smallest integer such that $d(y_j+1,i_j,x)<2^-j$.
I read there was use of first countable neighborhood basis in the comments, it is the same, you need to chose the next candidate of your sequence from $(y_k,l)$ in such a manner that you get 'closer' to $x$ each time.
In the space of tempered distributions with classical closure definition this statement is false since it is not sequential :
Suppose the statement is true, consider a familly $Omega_i$ that is constant, thent that statement for constant famillies of subspaces implies that the space is sequential therefore a contradiction.
answered Aug 3 at 16:13
jeanmfischer
478212
edited Aug 3 at 17:33
answered Aug 3 at 16:13
jeanmfischer
478212
answered Aug 3 at 16:13
jeanmfischer
478212
answered Aug 3 at 16:13
jeanmfischer
478212
478212
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
 |Â
show 8 more comments
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
(first countable implies sequential closure = topological clusure)
– jeanmfischer
Aug 3 at 16:22
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
Thank you very much for your comment. I more or less understand how to deal with first-countable spaces. The problem is that my main interest is space $S'(R^n)$ which is not 1st countable(math.stackexchange.com/questions/678785/…). I worked on that 3-4 hours and started to think that this is not true in that case. What is your intuition about that? Thanks in advance.
– redkin77
Aug 3 at 16:56
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
hi ! I just edited the awnser and by doing so awsered your question in your comment :)
– jeanmfischer
Aug 3 at 16:58
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
If the space is not even sequential i cannot build the double sequence and have no way of proving this since it would imply that the space is sequential taking the sequence of subspaces $Omega_i$ constant !
– jeanmfischer
Aug 3 at 17:01
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
Thanks for you feedback! Now I am looking for some hammer, some kind of substitute for "countable neighbourhood bases" in $S'(R^n)$? I think $S'(R^n)$ is too specific for that statement to be non-true.
– redkin77
Aug 3 at 17:04
 |Â
show 8 more comments
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You need the topological space to, at least, be sequential, otherwise you can pick a constant sequence $Omega_i=S$ where $S$ is a subspace that is sequentially closed but not closed. Then, any $xinoverline Ssetminus S$ will be in $bigcap_iinBbb Noverline Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_iinOmega_i$. That being said, it appears that spaces of tempered sequence always are.
– Saucy O'Path
Aug 3 at 14:58
Sorry, can you give an example of Hausdorf space, with such subset S?
– redkin77
Aug 3 at 15:06
Thank you very much for this comment! I think I understood what you mean. I did not even know that sequential closure could be different from the standard one. What if $overlineOmega$ is the sequential closure? When I was formulating I kept in mind sequential closure.
– redkin77
Aug 3 at 15:11
A quick example of non-sequential Hausdorff space is the set $X$ of Lebesgue-measurable functions from $(0,1)$ to $(0,1)$ endowed with the subspace topology inherited from the product space $(0,1)^(0,1)$. The function $int: Xto Bbb R,quad fmapsto int_0^1 f(x),dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $left fin X,:, 0<int_0^1 f(x),dx< 1right$ is not open and $int$ is not continuous.
– Saucy O'Path
Aug 3 at 15:53
It would be true if the space is first countable: then if $U_n$ is a countable neighborhood basis at $x$, then $U_1 cap cdots U_n cap Omega_n ne emptyset$ for each $n$, and choose $x_n$ in this set for each $n$.
– Daniel Schepler
Aug 3 at 15:55