Problem of the Week #8 - July 23rd, 2012

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Chris L T521

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Here is this week's problem!

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Problem: Let $X$ be a topological space. Show that $X$ is Hausdorff if and only if the diagonal $\Delta=\{x\times x:x\in X\}$ is closed in $X\times X$.

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Chris L T521

Well-known member
Staff member
This week's question was correctly answered by girdav. Here's my solution:

Proof: ($\Rightarrow$) Let $X$ be Hausdorff and let $\Delta=\{x\times x:x\in X\}$ denote the diagonal of $X\times X$. We seek to show that $\Delta$ is closed in $X\times X$. Suppose $x\times y\in X\times X\backslash\Delta$. Thus, $x\neq y$ and there exists disjoint open sets $U$ and $V$ such that $x\in U$ and $y\in V$. By the definition of the product topology on $X\times X$, $U\times V$ is an open subset of $X\times X$; furthermore, $U\times V\subset X\times X\backslash\Delta$ (otherwise, $U\cap V\neq\emptyset$ [contradicting the fact that $X$ is Hausdorff]). Since $U\times V$ is open in $X\times X\backslash\Delta$, $\Delta$ is closed in $X\times X$.

($\Leftarrow$) Let $\Delta=\{x\times x:x\in X\}$ be closed in $X\times X$. We seek to show that $X$ is Hausdorff. Again, consider $x\times y \in X\times X\backslash\Delta$, where $x\neq y$. Then there is a basis open set $U\times V\subset X\times X\backslash\Delta$ that contains $x\times y$ (in the product topology). Here, $U$ and $V$ are disjoint open sets containing $x$ and $y$ respectively. Therefore, $X$ is Hausdorff.

Q.E.D.

Here's girdav's solution:

Assume that the diagonal $\Delta$ is closed (for the product topology), and take $x,y\in X$ with $x\neq y$. Then $(x,y)\in X\times X\setminus\Delta$ which is open, hence we can find $U$ and $V$ two open subsets of $X$ such that $(x,y)\in U\times V\subset X\times X\setminus\Delta$. We have $x\in U$,$y\in V$ and if $z\in U\cap V$, $(z,z)$ wouldn't be in $\Delta$. Hence $U$ and $V$ are disjoint, which proves that $X$ is Hausdorff.

Conversely, assume that $X$ is Hausdorff. We shall show that the complement of the diagonal $C$ is open. Let $(x,y)\in C$. Then $x\neq y$ and we can find two disjoint open subsets of $X$, $U$ and $V$, such that $x\in U$ and $y\in V$. Then $U\times V$ is open for the product topology, and $U\times V\subset C$ (if $(x,y)\in\Delta$, $x=y$ and $(x,x)$ can't be in $U\times V$, since $U$ and $V$ are disjoint).

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