The product of Hausdorff spaces is Hausdorff

Question

I'm confused how it can be true that the product of an infinite number of Hausdorff spaces $X_\alpha$ can be Hausdorff.

If $\prod_{\alpha \in J} X_\alpha$ is a product space with product topology, the basis elements consists of of products $\prod_{\alpha \in J} U_{\alpha}$ where $U_{\alpha}$ would equal $X_{\alpha}$ for all but finitely many $\alpha$'s. If this is the case and we had two distinct points, $x$ and $y$, in $\prod_{\alpha \in J} X_{\alpha}$ and a basis element, $B_x$ containing $x$ then In looking for a basis element,$B_y$ that contains $y$ but is disjoint from $B_x$ then for every $\alpha$ such that the open set $U_\alpha$ of $B_x$ is equal to $X_\alpha$ the open set $U_\alpha$ for $B_y$ would have to be empty. But then it couldn't possibly contain $y$. (or any other point of $\prod_{\alpha \in J} X_\alpha$ for that matter). How then is it possible that $\prod_{\alpha \in J} X_\alpha$ is Hausdorff in the product topology?

What am I missing?

Two products $\prod_{\alpha\in J} U_\alpha$ and $\prod_{\alpha\in J}V_\alpha$ are disjoint when there is (at least) one $\alpha_0 \in J$ such that $U_{\alpha_0} \cap V_{\alpha_0} = \varnothing$. — Daniel Fischer, Sep 08 '13 at 19:01
Ah, I see. I'm so stupid I wasn't thinking about that right. Thank you. — TopologyQuestion, Sep 08 '13 at 19:05

score 52 · Answer 1 · edited Jan 26 '21 at 23:00

52

If $x$ and $y$ are distinct points of $\prod_{\alpha\in J}X_\alpha$, then there is at least one $\alpha_0\in J$ on which they differ, meaning that $x_{\alpha_0}\ne y_{\alpha_0}$. $X_{\alpha_0}$ is Hausdorff, so there are open sets $U_{\alpha_0}$ and $V_{\alpha_0}$ in $X_{\alpha_0}$ such that $x_{\alpha_0}\in U_{\alpha_0}$, $y_{\alpha_0}\in V_{\alpha_0}$, and $U_{\alpha_0}\cap V_{\alpha_0}=\varnothing$. Now let $U_\alpha=V_\alpha=X_\alpha$ for each $\alpha\in J\setminus\{\alpha_0\}$, let $U=\prod_{\alpha\in J}U_\alpha$, and let $V=\prod_{\alpha\in J}V_\alpha$; then $U$ and $V$ are basic open sets in $\prod_{\alpha\in J}X_\alpha$, $x\in U$, $y\in V$, and $U\cap V=\varnothing$. The only statement there that might not be immediately evident is that $U\cap V=\varnothing$; to see this, note that if $z\in U\cap V$, then $z_{\alpha_0}\in U_{\alpha_0}\cap V_{\alpha_0}=\varnothing$, so no such $z$ can exist.

Thus, $\prod_{\alpha\in J}X_\alpha$ is Hausdorff.

edited Jan 26 '21 at 23:00

MyMathYourMath

4,591

answered Sep 08 '13 at 19:05

Brian M. Scott

631,399

In the third line, do you mean $ U_{a_0} \cap V_{a_0}$ instead of $ U_{a_0} \cap V_{a_0}$? – Siyanda Nov 07 '13 at 09:36
@Siyanda: I mean $U_{\alpha_0}\cap V_{\alpha_0}$ instead of $U_{\alpha_0}\cap U_{\alpha_0}$. :-) Thanks; fixed. – Brian M. Scott Nov 07 '13 at 09:44
Oh, yes! Haha, that's what I meant too. :) – Siyanda Nov 07 '13 at 09:51
4

In the second and third lines, shouldn't it be $x_{\alpha_0}\in U_{\alpha_0}$ and $y_{\alpha_0}\in V_{\alpha_0}$? – soap Jan 17 '17 at 18:07
Can I ask a question, If we take the box and not the product topology does this still hold? – MyMathYourMath Jan 26 '21 at 22:56
@HossienSahebjame: Yes. If $\tau$ is the product topology, and $\tau_1$ is the box topology, then $\tau\subseteq\tau_1$, so the fact that $\tau$ is Hausdorff immediately implies that the finer topology $\tau_1$ is Hausdorff. – Brian M. Scott Jan 26 '21 at 22:58
one more thing, this uses the fact that: If $B \cap C = \emptyset$ then $(A \times B) \cap (A \times C) = \emptyset$? sorry I just want to know every single reason for every single move. – MyMathYourMath Jan 26 '21 at 23:01
1

@HossienSahebjame: Yes, it does. – Brian M. Scott Jan 26 '21 at 23:02
If $x$ and $y$ differ in an infinite number of indices, is it still true that $U$ and $V$ are basic open sets in the product topology? – R. H. Vellstra Oct 18 '22 at 21:02
@pyridoxal_trigeminus: yes: They differ on only one coordinate, $\alpha_0$. – Brian M. Scott Oct 19 '22 at 02:22

The product of Hausdorff spaces is Hausdorff

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