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Show that the topological space $Y$ is Hausdorff iff the diagonal $D = \{(y_1,y_2)\in Y\times Y: y_1=y_2\}$ is a closed subset of $Y\times Y$.

My attempt: First, suppose $Y$ is Hausdorff. Then $Y\times Y$ is Hausdorff, so for any $y\in Y\times Y - D$ and $y'\in D$, I can find open sets $U_y$ and $U_y'$ such that $U_y\cap U_y' = \varnothing$. This means that $y\in U_y\subseteq Y\times Y -D$, so $Y\times Y -D$ is open, which implies that $D$ is closed.

Does anyone see anything wrong with my argument?

For the converse, I'm a little stuck. My hunch is that since $D$ is closed, $Y\times Y-D$ is open, and I want to use this somehow to show that $Y\times Y$ is Hausdorff. Then we can conclude that $Y$ is Hausdorff. Any advice?

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user124910
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    I think your forward implication still needs some clarification, since we still haven't shown that $U_y \subset Y \times Y \setminus D$. However, if you instead take $V = U_y \setminus D$, then you should have the open set containing $y$ that you need for the rest of your proof. – Ryder Bergerud Feb 24 '14 at 08:08
  • Looking at this, I just realized I had another question. Wouldn't I have to say that (if I use $V_y=U_y-D$ as you defined above) $\bigcup_{y\in Y\times Y-D}V_y$ is open and equal to $Y\times Y-D$? – user124910 Feb 24 '14 at 08:19
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    See also http://math.stackexchange.com/questions/136922/x-is-hausdorff-if-and-only-if-the-diagonal-of-x-times-x-is-closed and the questions linked there. – Martin Sleziak Feb 24 '14 at 09:11

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Your argument is correct.

For the other implication, take $y_1,y_2$ in $Y$, with $y_1\neq y_2$. Your purpose is to separate this two points using open subsets of $Y$. But, $y_1\neq y_2$ means that the pair $(y_1,y_2)$ is not in $D$, thus it is in the complementary set $Y\times Y-D$, which is open. Thus you can find an open subset of $Y\times Y$, say $U$, such that $(y_1,y_2)\in U$ and $U\cap D=\varnothing$. By definition of product topology, you can factor $U$ as a cartesian product of two opens, say $U_1$ and $U_2$, with $y_1\in U_1$ and $y_2\in U_2$. The fact that $U=U_1\times U_2$ doesn't intersect the diagonal $D$ precisely means that $U_1$ and $U_2$ are disjoint.

Federica Maggioni
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