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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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
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    $\begingroup$ @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

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.

  • $\begingroup$ (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

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