If $X=\prod_{i \in I}X_i$ is regular, then $X_i$ is regular for every $i\in I$

Question

Let $\{X_i\}_{i \in I}$ be a family of non empty topological spaces and consider $X=\prod_{i \in I}X_i$. Show that if $X$ is regular, then $X_i$ is regular for every $i\in I$

Given $i \in I$ take $x_i \in X_{i}$ and $A \subseteq X_i$ a closed subset that doesn't contain $x_{i}$.

Consider $(y_j) \in X$ where $y_j = x_i$ if $j =i$ and $y_j$ is any element of $X_j$ if $j \neq i$.

Take $\prod A_i \subseteq X$ given by $A_j = \{y_j\}$ if $j \neq i$ and $A_j = A$ if $j=i$.

Since $X$ is regular by hipothesys, there exist open disjoint subsets $U, V \subseteq X$ such that $(y_j) \in U$, $\prod A_i \subseteq V$. Now since $y_j \in \pi_{j}(U)\cap\pi_{j}(V)$ for every $j \neq i$ we have that $\pi_{i}(U) \cap \pi_{i}(V) = \emptyset$.

However, nothing assures me that $\pi_{i}(V)$ is open in $X_i$ and if I take basic element of $X$ contained in $V$, nothing assures me that the $i$-coordinate of the basic element contains $A$ entirely.

So I took, a basic element $\prod U_j \subseteq U$ that contains $(y_j)$ and necessairily, $U_i \neq X_i$, since $U_i \subseteq \pi_{i}(U)$ and $\pi_{i}(U) \cap \pi_{i}(V) = \emptyset$.

So my idea is to take my open subset containing $A$ in $X_i$ as $\overline{U_i}^c$ suggested by the fact that $U_i \cap A = \emptyset$ but I wasn't able to prove that $\overline{U_i}\cap A =\emptyset$ (which in fact, im not sure if it is even true)

Am I on the correct track? Any other hint?

Answer 1

The system is unhappy with the number of comments we have been adding so I will add a (hopefully final) comment as an answer.

It seems that we have reduced the problem of whether or not your $\pi_{i}(U)$ and $\pi_{i}(V)$ are the requisite open sets to the question of whether or not every nonempty open set in $\prod X_{j}$ is of the form $\prod U_{j}$ where $U_{j}=X_{j}$ for all but finitely many $j$. These are certainly basic open sets.

Then let $O\subseteq X=\prod X_{j}$ be an open set. Then $O=\bigcup_{\alpha\in J} B_{\alpha}$ where the $B_{\alpha}$ are any basic open sets. We claim that $O$ itself is a basic open set. To see this we note that if $B_{\alpha}$ and $B_{\beta}$ are given with $\alpha\neq\beta$ then there is a finite set $I_{\alpha}=\{j_{\alpha_{1}},\ldots, j_{\alpha_{n}}\}$ such that $\pi_{j_{\alpha_{k}}}(U)\neq X_{j_{\alpha_{k}}}$ (please forgive this hurricane of indices). Likewise there is a finite set $I_{\beta}=\{j_{\beta_{1}},\ldots, j_{\beta_{m}}\}$ with similar characteristics. Then we consider $B_{\alpha}\cup B_{\beta}$. If $I_{\alpha}\cap I_{\beta}=\emptyset$ then $B_{\alpha}\cup B_{\beta}=X$. Otherwise $B_{\alpha}\cup B_{\beta}$ is a basic open set $B_{\gamma}$ where $I_{\gamma}=I_{\alpha}\cap I_{\beta}$.

In general, if $\pi_{j}(O)\neq X_{j}$ for infinitely $j$ then define $J_{O}$ to be the set of indices $j$ for which $\pi_{j}(O)\neq X_{j}$. We then have that, for all $\alpha\in J$, $\pi_{j}(B_{\alpha})\neq X_{j}$ for all $j\in J_{O}$. However, this implies that infinitely components of $B_{\alpha}$ are not equal to an $X_{j}$, which is a contradiction. Therefore $O$ must be a basic open set. In particular $O$ is the basic open set whose $i^{th}$ component is $\bigcup_{\alpha\in J}\pi_{j}(B_{\alpha})$.

${\bf Note:}$ This was not terribly rigorous. If you wanted to make it so you could use transfinite induction on the basic open sets that make up the open set $O$.

Answer 2

let $X=\prod{X_{\alpha}}$ be a regular space $\implies$ X is Hausdorff space $\implies$ $X_{\alpha}$ is a Hausdorff space for all $\alpha\implies$ one point sets in $X_{\alpha}$ are closed set
now consider space $X_{\beta}$ let $x_{\beta}$ be any point of $X_{\beta}$ and $A_{\beta}$ be any closed set in $X_{\beta}$ such that $x_{\beta}\not\in A_{\beta}$
now for each $\alpha$ different from $\beta$ choose $x_{\alpha}\in X_{\alpha}$ set $$A =\prod{A_{\alpha}}$$ where $A_{\alpha}=\{x_{\alpha}\}$ if $ \alpha \neq \beta$ now A is closed in X (why?) let $x\in X$ be such that $ \pi_{\alpha}(x) = x_{\alpha}$ where $\pi_{\alpha}$ is $\alpha$ coordinate map, as $\pi_{\beta}(x)=x_{\beta}\not\in A_{\beta} \implies x\not\in A$
then by regularity of X choose set U and V open in X such that $$x\in U , A\subset V$$ and $$U\cap V =\emptyset$$
now define $B=\prod{B_{\alpha}}$ where $B_{\alpha}= \{x_{\alpha}\}$ if $\alpha\neq\beta$ and $B_{\beta}=X_{\beta}$
give B the subspace topology, here we make following observations \

$$x\in U\cap B$$ $$A\subset V\cap B$$ let f be the restriction of $\pi_{\beta}$ to B, it is easy to see that f is a homeomorphism set $U_{0}=f(U\cap B) $ and $V_{0}=f(V\cap B) $
as $U\cap B$ and $V\cap B$are disjoint and open and f is a homeomorphism $\implies$ $U_{0}$ and $V_{0}$ are disjoint and open, also $$f(x)=x_{\beta}\in U_{0}$$ $$f(A)=A_{\beta}\subset V_{0}$$