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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.