2
$\begingroup$

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 $\overline{\Omega}$ we denote a sequential closure of $\Omega\subseteq H$. Is it true that:

$x\in \cap_{i=1}^\infty \overline{\Omega_i}$ if and only if there is sequence $\{x_i\}$ such that $x_i \rightarrow x, x_i\in \Omega_i$.

The fact that from $x_i \rightarrow x, x_i\in \Omega_i$ we can deduce $x\in \cap_{i=1}^\infty \overline{\Omega_i}$ is obvious. The opposite is problematic.

I tried to prove the opposite statement via reducing to Cantor's intersection theorem. Suppose that $x\in \cap_{i=1}^\infty \overline{\Omega_i}$ is fixed. Then I define $R_i = \{\{x_n\}| \exists N, \forall n>N: x_n \in \Omega_i, x_n \rightarrow x\}$ (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: $R_i\supseteq R_{i+1}$.

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.

$\endgroup$
  • $\begingroup$ 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 $x\in\overline S\setminus S$ will be in $\bigcap_{i\in\Bbb N}\overline \Omega_i$ but it won't be the limit of any sequence $x_i$ such that $x_i\in\Omega_i$. That being said, it appears that spaces of tempered sequence always are. $\endgroup$ – Saucy O'Path Aug 3 '18 at 14:58
  • $\begingroup$ Sorry, can you give an example of Hausdorf space, with such subset S? $\endgroup$ – redkin77 Aug 3 '18 at 15:06
  • $\begingroup$ 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 $\overline{\Omega}$ is the sequential closure? When I was formulating I kept in mind sequential closure. $\endgroup$ – redkin77 Aug 3 '18 at 15:11
  • $\begingroup$ 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 $\int: X\to \Bbb R,\quad f\mapsto \int_0^1 f(x)\,dx$ is sequentially continuous by dominated convergence theorem, but it is surjective on all non-empty open sets. Therefore, $\left\{ f\in X\,:\, 0<\int_0^1 f(x)\,dx< 1\right\}$ is not open and $\int$ is not continuous. $\endgroup$ – Saucy O'Path Aug 3 '18 at 15:53
  • $\begingroup$ 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 $n$. $\endgroup$ – Daniel Schepler Aug 3 '18 at 15:55
2
$\begingroup$

For a counterexample using sequential closures, consider the topological space whose underlying set is $\mathbb{N}^2 \sqcup \{ x_0 \}$, and with the topology such that $U$ is open if and only if $x_0 \notin U$ or for some function $f : \mathbb{N} \to \mathbb{N}$, $\{ (x, y) \in \mathbb{N}^2 \mid y > f(x) \} \subseteq U$.

Now, let $\Omega_n := \{ (x, y) \in \mathbb{N}^2 \mid x \ge n \}$. 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 $f : \mathbb{N} \to \mathbb{N}$ such that $y_n < f(x_n)$ for each $n$. It follows that $(x_n, y_n) \not\to x_0$ as $n \to \infty$ since the corresponding neighborhood of $x_0$ for this $f$ does not contain any element of the sequence.

$\endgroup$
  • $\begingroup$ Nice counterexample, I am so jealous of it :) $\endgroup$ – jeanmfischer Aug 3 '18 at 17:59
  • 1
    $\begingroup$ Obviously heavily inspired by the prototypical counterexample showing that sequential closure is not idempotent in general... $\endgroup$ – Daniel Schepler Aug 3 '18 at 18:00
  • $\begingroup$ 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. $\endgroup$ – redkin77 Aug 3 '18 at 18:04
  • $\begingroup$ Is this the space where non idempotency is realised ? $\endgroup$ – jeanmfischer Aug 3 '18 at 18:46
  • 1
    $\begingroup$ @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 $n\to \infty$ and $y_m \to z$ as $m\to \infty$ but no sequence within $\{ x_{mn} \}$ converges to $z$. Then my example here is essentially the quotient identifying all $y_m$ and $z$. $\endgroup$ – Daniel Schepler Aug 3 '18 at 18:57
1
$\begingroup$

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 $x \in \cap_i \bar{\Omega_i}$, then for each $i$, there is a sequence $(y_{i,n})_n$ in $\Omega_i$ that converges to $x$, since $x \in \bar{\Omega_i}$. 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.

$\endgroup$
  • $\begingroup$ (first countable implies sequential closure = topological clusure) $\endgroup$ – jeanmfischer Aug 3 '18 at 16:22
  • $\begingroup$ 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. $\endgroup$ – redkin77 Aug 3 '18 at 16:56
  • $\begingroup$ hi ! I just edited the awnser and by doing so awsered your question in your comment :) $\endgroup$ – jeanmfischer Aug 3 '18 at 16:58
  • $\begingroup$ 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 ! $\endgroup$ – jeanmfischer Aug 3 '18 at 17:01
  • $\begingroup$ 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. $\endgroup$ – redkin77 Aug 3 '18 at 17:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.