Is the product space Hausdorff if both $X$ and $Y$ are Hausdorff?

[Chris L T521]

Here's this week's problem.

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Problem: Let $X$ and $Y$ be two non-empty topological spaces. Show that the product space $X\times Y$ is Hausdorff if and only if $X$ and $Y$ are Hausdorff.

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[Chris L T521]

Sorry about posting this late guys! Some things happened last night and it slipped my mind; I was going to do it this morning, but then I overslept... >_>.

No one answered this week's question. Here's my solution below.

Spoiler

Proof: Suppose $X$ and $Y$ are Hausdorff. Then for any $x_1,x_2\in X$, $y_1,y_2\in Y$, there are open neighborhoods $U_1\ni x_1,U_2\ni x_2 \subset X$ and $V_1\ni y_1,V_2\ni y_2\in Y$ such that $U_1\cap U_2=\emptyset$ and $V_1\cap V_2=\emptyset$. Now, consider the product space $X\times Y$. Our open sets are of the form $U_i\times V_i$. So in this case consider the sets $U_1\times V_1$ and $U_2\times V_2$. We now see that $(U_1\times V_1)\cap(U_2\times V_2) = (U_1\cap U_2)\times (V_1\cap V_2)$. Since $X$ and $Y$ are Hausdorff, we get $(U_1\times V_1)\cap (U_2\times V_2) = \emptyset$. Therefore, $X\times Y$ is Hausdoff.

Conversely, suppose $X\times Y$ is Hausdorff. Then for any two points $x_1\times y_1,x_2\times y_2\in X\times Y$, there exist open neighborhoods $U_1\times V_1\ni x_1\times y_1, U_2\times V_2\ni x_2\times y_2\subset X\times Y$ such that $(U_1\times V_1)\cap(U_2\times V_2)=\emptyset$. However, as seen above, this is equivalent to $(U_1\cap U_2)\times (V_1\cap V_2)=\emptyset$. Thus the only way to have this equality is if we have $U_1\cap U_2=\emptyset$ and $V_1\cap V_2=\emptyset$. Therefore, $X$ and $Y$ are Hausdorff.

Q.E.D.