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I need help with the following proof. Would be thankful for any help.

If $X, Y$ are nonempty topological spaces. Given the $X \times Y$ is Hausdorff, show that $X$ and $Y$ each are Hausdorff spaces.

I tried to look at given points $x_1, x_2$ in $X$ with $x_1 \ne x_2$ and $y_1, y_2$ in $Y$ with $y_1 \ne y_2$. Then I used the definition of the product topology and the Hausdorff property of $X \times Y$, to find open neighborhoods $U_1 \times V_1$ of $(x_1, y_1)$ and $U_2 \times V_2$ of $(x_2, y_2)$, $U_1, U_2$ open in $X$ and $V_1, V_2$ open in $Y$. Then I used the Hausdorff property to say that the intersection of $U_1 \times V_1$ and $U_2 \times V_2$ is empty. This also means $(U_1 \cap U_2) \times (V_1 \cap V_2)$ is empty. This means either $U_1 \cap U_2$ or $V_1 \cap V_2$ is empty. I can't really figure out how to prove that if $U_1 \cap U_2$ is empty and $V_1 \cap V_2$ is not empty, that $Y$ is Hausdorff.

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    $\begingroup$ What have you tried? Where are you stuck? $\endgroup$
    – K. Jiang
    Apr 18 at 17:19
  • $\begingroup$ I tried to look at given points x1,x2 in X with x1≠x2 and y1,y2 in Y with y1≠y2. Then I used the definition of the producttopology and and the Hausdorff property of X x Y, to find open neighboorhoods U1 x V1 of (x1,y1) and U2 x V2 of (x2,y2) U1,U2 open in X and V1,V2 in Y. Then I used the Hausdorff property to say the the intersection of (U1 x V1) and (U2 x V2) is empty. This is also means (U1 n U2) x (V1 n V2) is empty, n being the intersection symbol. This means either U1 n U2 or V1 n V2 is empty. I cant really figure how to prove that if U1 n U2 is empty and V1 n V2 is not empty, that Y $\endgroup$
    – alex.
    Apr 18 at 17:32
  • $\begingroup$ is Hausdorff. Or for the analog case. $\endgroup$
    – alex.
    Apr 18 at 17:33
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    $\begingroup$ You should include those details in the body of your question, so that your question is better-received. $\endgroup$
    – K. Jiang
    Apr 18 at 17:33
  • $\begingroup$ oh yes my bad, im so stuck on this problem, that i completely forgot. $\endgroup$
    – alex.
    Apr 18 at 17:35

1 Answer 1

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Let $x_1 \ne x_2 \in X$. Consider any $y \in Y$. Since $X \times Y$ is Hausdorff, there exist disjoint basis elements $U_1 \times V_1, U_2 \times V_2$ of the product topology such that $x_1 \times y \in U_1 \times V_1$ and $x_2 \times y \in U_2 \times V_2$. By definition, $U_1, U_2$ are open in $X$ and $V_1, V_2$ are open in $Y$. We claim that $U_1$ and $U_2$ are disjoint. Suppose not. Then there exists $x_0 \in U_1 \cap U_2$, and $x_0 \times y \in (U_1 \times V_1) \cap (U_2 \times V_2)$, contrary to the hypothesis that $U_1 \times V_1$ and $U_2 \times V_2$ are disjoint. Since $x_1 \ne x_2$ were arbitrary, $X$ is Hausdorff.

A very similar argument applies to show that $Y$ is also Hausdorff.

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  • $\begingroup$ this makes alot of sense, thank you for your help. could one also prove in the way i tried to, or is the approach insensible? $\endgroup$
    – alex.
    Apr 18 at 17:38
  • $\begingroup$ Yes, you have the right idea. A few comments: a neighborhood of the point $x \times y$ in $X \times Y$ under the product topology need not be of the form $U \times V$, where $U$ is open in $X$ and $V$ open in $Y$. However, since these sets form a basis for the product topology, Hausdorffness implies the existence of disjoint basis elements containing unequal points (you should prove this if you have not seen it). Then, it is true that $(U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \times V_2)$. Since $V_1 \times V_2 \ne \emptyset$, then $U_1 \cap U_2 = \emptyset$. $\endgroup$
    – K. Jiang
    Apr 18 at 17:43
  • $\begingroup$ oh yes i just wrote it down how my professor does in his lectures, i should have mentioned that U x V form a basis for the product topology. Im not sure if I understood the last part of your comment, could you elaborate on that? $\endgroup$
    – alex.
    Apr 18 at 17:53
  • $\begingroup$ I made a typo; it should be $V_1 \cap V_2 \neq \emptyset$ rather than $V_1 \times V_2 \neq \emptyset$ (which is nonsensical). According to my choice of points $x_1 \times y$ and $x_2 \times y$, and disjoint basis elements $U_1 \times V_1$ and $U_2 \times V_2$ containing them, then $y \in V_1 \cap V_2 \implies V_1 \cap V_2 \neq \emptyset$. Since $(U_1 \times V_1) \cap (U_2 \times V_2) = (U_1 \cap U_2) \times (V_1 \cap V_2) = \emptyset$, and $V_1 \cap V_2 \neq \emptyset$, then it must be the case that $U_1 \cap U_2 = \emptyset$, which is what want to show. $\endgroup$
    – K. Jiang
    Apr 18 at 18:11
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    $\begingroup$ Alright I figured so, but was not sure if it was the case. I understand what you mean. I guess I should have tried to prove Hausdorffness for X and Y individually. Or rather I don‘t know how to do it, if I dont assume y in Y as an arbitrary fixed point. Nevertheless thank you so much for your help, it was really insightful. $\endgroup$
    – alex.
    Apr 18 at 18:20

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