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By preserved by products I mean - $\prod X_{\alpha}$ has property $P$ iff $X_{\alpha}$ has property $P$ for all $\alpha$ in index set

Also, $X$ is Completely Hausdorff if for $x\neq y$ in $X$, $\exists$ continuous function $f:X\to I$ with $f(x) = 0$, $f(y) = 1$.

Are Semiregularity and Completely Hausdorff preserved by products? If not, then is any direction of the iff statement above true?

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  • $\begingroup$ Is there supposed to be an implicit hypothesis that all the $X_\alpha$-s should be non-empty? $\endgroup$
    – user239203
    Jul 19 '20 at 23:13
  • $\begingroup$ @Gae.S.: Yes, but it doesn’t matter, since the empty space is vacuously semiregular and completely Hausdorff. $\endgroup$ Jul 19 '20 at 23:19
  • $\begingroup$ @BrianM.Scott It does matter for that very reason, since there's an if and only if. $\endgroup$
    – user239203
    Jul 19 '20 at 23:20
  • $\begingroup$ @Gae.S.: Ah, I was looking at the title question and missed the alteration in the body. $\endgroup$ Jul 19 '20 at 23:21
  • $\begingroup$ Yes, all spaces are assumed non empty $\endgroup$
    – Ishan Deo
    Jul 20 '20 at 6:34
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Suppose that $X_\alpha$ is completely Hausdorff for each $\alpha\in A$, and let $x,y\in X$ be distinct points; there is an $\alpha\in A$ such that $x_{\alpha}\ne y_{\alpha}$, and there is continuous $f_\alpha:X_\alpha\to[0,1]$ such that $f_\alpha(x_\alpha)=0$ and $f_\alpha(y_\alpha)=1$. Now define

$$f:X\to[0,1]:z\mapsto f_\alpha(z_\alpha)\;;$$

if $\pi_\alpha:X\to X_\alpha$ is the projection map, $f=f_\alpha\circ\pi_\alpha$. Clearly $f$ is continuous, $f(x)=0$, and $f(y)=1$. Thus, $X$ is completely Hausdorff.

Conversely, if $X$ is completely Hausdorff and non-empty, then each $X_\alpha$ is completely Hausdorff: complete Hausdorffness is evidently hereditary, and if we fix $x\in X$, the subset

$$\big\{y\in X:y_\beta=x_\beta\text{ for all }\beta\in A\setminus\{\alpha\}\big\}$$

of $X$ is homeomorphic to $X_\alpha$.

Now suppose that each $X_\alpha$ is semiregular, and let $\mathscr{B}_\alpha$ be a base of regular open sets for $X_\alpha$. Then $X$ has a base $\mathscr{B}$ whose elements are the sets, $\prod_{\alpha\in A}U_\alpha$ such that $U_\alpha=X_\alpha$ for all but finitely many $\alpha\in A$, and $U_\alpha\in\mathscr{B}_\alpha$ whenever $U_\alpha\ne X_\alpha$. Let $B=\prod_{\alpha\in A}U_\alpha\in\mathscr{B}$, and let $F=\{\alpha\in A:U_\alpha\ne X_\alpha\}$. It’s easy to check that the sets $\pi_\alpha^{-1}[U_\alpha]$ for $\alpha\in F$ are regular open in $X$; $B=\bigcap_{\alpha\in F}\pi_\alpha^{-1}[U_\alpha]$, and the intersection of finitely many regular open sets is regular open, so $B$ is regular open, and $X$ is semiregular.

I am not at the moment sure about the other direction, since semiregularity is not hereditary.

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  • $\begingroup$ When you write $f=f_\alpha\circ\pi_\alpha$, this holds for all $\alpha$, right? Or just the $\alpha$ for which $x_\alpha \neq y_\alpha$? $\endgroup$
    – Ishan Deo
    Jul 20 '20 at 8:11
  • $\begingroup$ @IshanDeo: Just the $\alpha$ for which $x_\alpha\ne y_\alpha$: we use $\pi_\alpha$ to project the product space to the factor $X_\alpha$, and then we apply the function $f_\alpha:X_\alpha\to[0,1]$. $\endgroup$ Jul 20 '20 at 17:17
  • $\begingroup$ Can we say anything about the other $\alpha$? $\endgroup$
    – Ishan Deo
    Jul 20 '20 at 19:50
  • $\begingroup$ @IshanDeo: There’s no reason to say anything about them. We want a function from $X$ to $[0,1]$ that separates $x$ and $y$, and $f_\alpha\circ\pi_\alpha$ does the job. $\endgroup$ Jul 20 '20 at 19:51
  • $\begingroup$ I know there's no need. I was just wondering if there's possibly anything we could say about them. $\endgroup$
    – Ishan Deo
    Jul 20 '20 at 22:04
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Adding on to the answer by Brian above, for future reference if anyone is interested -

If $\prod X_\alpha$ is semiregular, $X_\alpha$ is indeed semiregular for all $\alpha$. This is proved in -

Porter, Grant Woods: Extensions and Absolutes of Hausdorff Spaces

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