# Are Semiregularity and Complete Hausdorff Properties preserved by Products

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?

• Is there supposed to be an implicit hypothesis that all the $X_\alpha$-s should be non-empty?
– user239203
Jul 19 '20 at 23:13
• @Gae.S.: Yes, but it doesn’t matter, since the empty space is vacuously semiregular and completely Hausdorff. Jul 19 '20 at 23:19
• @BrianM.Scott It does matter for that very reason, since there's an if and only if.
– user239203
Jul 19 '20 at 23:20
• @Gae.S.: Ah, I was looking at the title question and missed the alteration in the body. Jul 19 '20 at 23:21
• Yes, all spaces are assumed non empty Jul 20 '20 at 6:34

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.

• When you write $f=f_\alpha\circ\pi_\alpha$, this holds for all $\alpha$, right? Or just the $\alpha$ for which $x_\alpha \neq y_\alpha$? Jul 20 '20 at 8:11
• @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 $f_\alpha:X_\alpha\to[0,1]$. Jul 20 '20 at 17:17
• Can we say anything about the other $\alpha$? Jul 20 '20 at 19:50
• @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. Jul 20 '20 at 19:51
• I know there's no need. I was just wondering if there's possibly anything we could say about them. Jul 20 '20 at 22:04

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