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Given a product $X=\prod_{k\in K}X_k$ of topological spaces, is one or both of the following true?

$\operatorname{cl}(\pi_k^{-1}(U_k)) \subseteq \pi_k^{-1}(\operatorname{cl}(U_k))\quad $ or $\quad \operatorname{cl}(\pi_k^{-1}(U_k)) = \pi_k^{-1}(\operatorname{cl}(U_k))$ .

$U_k$ is an open set in $X_k$.

Mohamed Ali
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    Your question is essentially about the product of just two spaces. Just consider $ X_k $ and the product of the remaining factors as one "big" factor. Thus, @KaboMurphy was right to talk about two factors only. – Wlod AA Nov 04 '19 at 10:11
  • Below, please find my limitless proof :). It doesn't assume any openness. (I hope that I didn't mess up. Certainly, the general statement -- without openness -- is true regardless of the quality of my editing). – Wlod AA Nov 04 '19 at 11:20

4 Answers4

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I will write the proof when we have a product of two spaces and $\pi_k$ is the first projection but the proof is similar in the general case. Let $(x,y) \in cl(\pi_1^{-1}(U_1))$. Then we can write $(x,y)=\lim (x_i,y_i) $ with $x_i= \pi_k(x_i,y_i) \in U_k$. Now $\pi_1(x,y) =\lim \pi_1(x_i,y_i) \in cl(U_1)$ proving the first inclusion.

Now let $(x,y) \in \pi_1^{-1} (cl(U_1))$. Then $x \in cl(U_1)$ so we can write $x =\lim x_i$ with $x_i=\pi_1(x_i,y_i) \in U_1$ for all $i$. Now $(x,y)=\lim (x_i,y) \in cl(\pi_1^{-1}(U_1))$. This proves the second part.

Note: we cannot always use sequences. In general we have to use nets in this proof.

Note that a point $x$ in the closure of a set $U$ iff there is a net in $U$ converging to $x$.

Proof of second part without using nets: Let $(x,y)$ belong to RHS. If possible, suppose $(x,y) \notin$ LHS. Then there is an open set $V$ containing $(x,y)$ such that $V \cap \pi_1^{-1}(U_1)=\emptyset$. Let $V_0=\{z:(z,y) \in V\}$. Then $V_0$ is open because $z \to (z,y)$ is continuous. Also $x \in V_0$ Since $x \in cl(U_1)$ there exists some point $z$ in $V_0\cap U_1$. Now $(z,y) \in V \cap \pi^{-1}(U_1)=\emptyset$, a contradiction.

Kavi Rama Murthy
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  • You need to define lim in the cases of general spaces (sure, use Bourbaki, but...). Then you need to quote the respective theorems. This is not a clean way to do it but rather cumbersome, I'de think. – Wlod AA Nov 04 '19 at 09:43
  • @WlodAA My proof is accessible to those who. are familiar with nets, but it makes full sense to others when the spaces are metrizable. I have added a note saying that I am using nets, not sequences. – Kavi Rama Murthy Nov 04 '19 at 09:51
  • Kabo, nets are fine but not elegant. I feel bad that I used them at one time in the past. I was always presenting that theorem later without nets nor any limits. Bourbaki's approach to limits is clearly nicer. And still, it is not necessary. Perhaps, after I get some sleep and find a moment, I'll extend my naked answer above to present a proof free of any limits. (However, the Bourbaki's proof of the Tikhonov's theorem using limits of ultrafilters is super easy!). – Wlod AA Nov 04 '19 at 10:00
  • To be honest I personally wouldn't yet understand the proof using nets. But on the other side I want to be formal at some level. If I understand correctly $(x,y)=\lim(x_i,y_i)$ iff for every open neighborhood $U$ of $(x,y)$ there is $N$ such that $i>N\to (x_i,y_i)\in U$. But how do I know that there is such a sequence? – Mohamed Ali Nov 04 '19 at 10:01
  • @WlodAA That would be welcomed! – Mohamed Ali Nov 04 '19 at 10:02
  • @AaronLenz, I truly intend to do it (I am simply old with all the consequences of this). Also, I would replace the Bourbaki's filters (which are nicer than nets) by something still more elegant, even if basically quite similar. My approach to this topic is purely topological. Somehow, ths was not important enough to do, and still too bad. – Wlod AA Nov 04 '19 at 10:08
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    @AaronLenz I have given a proof of the second part without using nets. – Kavi Rama Murthy Nov 04 '19 at 10:20
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In general for product spaces for $A_i \subseteq X_i$ for all $i$:

$$\operatorname{cl}\left(\prod_{i \in I} A_i\right) = \prod_{i \in I} \operatorname{cl} A_i$$

(See my answer here for a proof.) And your statement (the $=$ version) is just a special case where all $A_i=X_i$ except for $i=k$, where you have $A_k=U_k$. Openness is irrelevant.

Henno Brandsma
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The first one is clearly true because $\pi_k^{-1}({\rm cl}(U_k))$ is a closed set containing $\pi_k^{-1}(U_k)$

Gribouillis
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  • Very nice simple argument. I need only the inequality for my purpose now, but I added the equality for completeness and with some skepticism. – Mohamed Ali Nov 04 '19 at 10:06
  • @AaronLenz, I feel that your scepticism was unwarranted. Furthermore, we don't even need openness -- arbitrary subset will do. – Wlod AA Nov 04 '19 at 10:14
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Let $(X\,\, S)$ and $(Y\,\, T)$ be two topological spaces. Let $B\subseteq Y$ be arbitrary (not necessarily open). I'll show the a bit harder inclusion, that

$$ c(q(B))\supseteq q(c(B)) $$

where $c$ is the closure operation, and $q$ operates on subsets of $Y$ as the inverse of the projection $X\times Y\rightarrow Y.$

Proof: $$(x\,\,y)\in q(c(B))\,\, \Leftarrow:\Rightarrow\,\, y\in c(B)\,\,\Leftarrow:\Rightarrow $$

$$ \forall_{V\in T} (y\in V\Rightarrow V\cap B\ne\emptyset)\,\, \Leftarrow:\Rightarrow $$

$$ \forall_{U\in S}\forall_{V\in T}\,((x\in U)\wedge(y\in V))\Rightarrow (U\times V)\cap(X\times B)\,\ne\emptyset\,\,\Leftarrow:\Rightarrow $$ $$ (x\,\,y)\in c(q(B)) $$

Actually, I have proved more, namely the equality

$$ c(q(B)) = c(q(B)) $$

Great!

Wlod AA
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