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Feedback on precision of argument and writing style is much needed.

Suppose that $(X, \mathcal{T})$ is a topological space. Show that if $X$ is Hausdorff, the diagonal $\Delta = \{(x,x) \mid x \in X\}$ is closed in $X \times X$.

ANSWER: Suppose $X$ is Hausdorff. We have to show that $\Delta$ is closed in $X \times X$, it is adequate to show that every point of closure of $\Delta = \{(x,x) \mid x \in X\}$ belongs to $\Delta$. Let $\delta = (u,v)$ be any point of closure of $\Delta$. We have to show that $(u,v) \in \Delta$, i.e $u=v$. Since $\delta = (u,v)$ is a point of closure of $\Delta$, thus for every open neighbourhood $W$ of $u$ and $V$ of $v$, $W(u) \times V (v)\cap \Delta \neq \emptyset$ . Consequently, $W(u) \cap V(v) \neq \emptyset$ for all open neighbourhoods $W,V$ of $u$ and $v$. However since $X$ is Hausdorff, $u$ and $v$ cannot be distinct, and so $u=v$.

How can I prove the converse?

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marked as duplicate by Najib Idrissi, kingW3, Jack D'Aurizio, N. F. Taussig, Dario Feb 21 '15 at 14:33

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    $\begingroup$ Since you try to mention every tiny step in the proof, you should mention explicitly what you used about the topology of $X\times X$: The sets of type $W\times V$ form a basis (and only this lets you find $W(u)\times V(v)\cap \Delta=\emptyset$). - Otherwise I think this is fine, though I personally would have shown $X\times X\setminus \Delta$ is open instead. $\endgroup$ – Hagen von Eitzen Feb 21 '13 at 16:53
  • $\begingroup$ I actually did this proof without having read up on basis or product topologies, so could you please elaborate? $\endgroup$ – Fatsho Feb 21 '13 at 16:56
  • $\begingroup$ @Fatsho You showed a set was closed in a topology without knowing how that topology is defined? Very impressive. $\endgroup$ – JSchlather Feb 21 '13 at 17:13
  • $\begingroup$ @JacobSchlather: Do you suppose that's why I am getting stuck with the converse? Because I'm clueless about product topologies. $\endgroup$ – Fatsho Feb 21 '13 at 17:17
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$\Delta$ is closed

$\iff$ $X\times X\setminus \Delta$ is open

$\iff$ For any $(u,v)\notin \Delta$ there is an open neighbourhood $W$ disjoint to $\Delta$

$\iff$ For any $(u,v)\notin \Delta$ there is an open base set $W=U\times V$ disjoint to $\Delta$, where $U,V$ are open neighbourhoods of $u$ and $v$, respectively.

$\iff$ For any $u,v\in X$ with $u\ne v$ there are neighbourhoods $U\ni u,V\ni v$ such that there is no $z\in X$ with $(z,z)\in U\times V$

$\iff$ For any $u,v\in X$ with $u\ne v$ there are neighbourhoods $U\ni u,V\ni v$ such that $U\cap V=\emptyset$

$\iff$ $X$ is Hausdorff.

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First, by definition, the topology on $X \times X$ is the one generated by the base $\mathcal{B} = \left\{ U \times V: U,V \mbox{ open in } X \right\}$. One can easily check that this is indeed a base for a topology, and it is the smallest topology that makes both the projection maps (onto the first and second coordinate) continuous. In general, for an arbitrary product, this is also how the topology is defined (the smallest one to make all projections continuous).

Now, suppose $\Delta$ is closed in $X \times X$. We want to show that $X$ is Hausdorff, so we take any two distinct points $x \neq y$, both in $X$. Then $(x,y)$ is by definition not in $\Delta$ and so, as $\Delta$ is closed, there is a basic open set $U \times V \in \mathcal{B}$ that contains $(x,y)$ and is disjoint from $\Delta$.

Now, $(x,y) \in U \times V$ iff $x \in U$, $y \in V$ and $U,V$ are both open. And they are disjoint, because if $z$ were in both of them, $(z,z) \in \Delta \cap U \times V$, which is a contradiction with the way we chose $U \times V$. So $X$ is Hausdorff.

The reverse implication is essentially the same idea: if $X$ is Hausdorff, then we can show that $\Delta$ is closed by showing any point not in $\Delta$, so a point $(x,y)$ with $x \neq y$, has a neighbourhood that misses $\Delta$ (so it's not in the closure of $\Delta$). Then of course we take disjoint $U$ and $V$, open around $x$ resp. $y$ and of course $U \times V$ is the required neighbourhood again...

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I assume that $X\times X$ is endowed with the product topology. Suppose $\Delta$ is closed, and assume to the contrary that $X$ is not Hausdorff. There exist $a,b \in X$ such that for every open neighborhoods $U$ of $a$ and $V$ of $b$ we have $U \cap V \neq \emptyset$. Therefore, every basic neighborhood $U\times V$ of $(a,b)$ in $X\times X$ (with the product topology) satisfies $(U\times V)\cap \Delta \neq \emptyset$. That is $(a,b)\in \bar{\Delta}$, and since $(a,b)\notin \Delta$ we have a contradiction to the fact that $\Delta$ is closed.

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