Proving a certain implication in the equivalent formulations of Hausdorff spaces

Question

Given on Wikipedia here, we see these are equivalent for a topological space $X$:

1. $X$ is Hausdorff
2. Limits of nets in $X$ are unique
3. Limits of filters on $X$ are unique.
4. Let $U_x$ denote an open neighborhood of $x \in X$, and let $$\mathcal X := \left\{ \overline{U_x} \; \middle| \;U_x \text{ is a neighborhood of } x \right\}$$ Then $\{x\} = \bigcap \cal X$ $\forall x \in X$.
5. The diagonal relation set $\Delta := \left\{ (x,x) \;\middle|\; x \in X \right\}$ is a closed set in $X^2$.

I am trying to prove the equivalence of some of these. In particular, I'm stuck on showing that $(4) \!\! \implies \!\! (5)$. (And I'd like to prove it as directly as possible, rather than circumventing through the other equivalent formulations. I have seen several proofs of, say, $(1) \!\!\implies \!\!(4)$ or $(1)\!\! \implies \!\!(5)$ here on MSE and elsewhere, but I'd rather go for $(4) \!\!\implies \!\!(5)$, so they've not been too helpful for me.)

I know that it is sufficient to show that $\Delta^C$ (the complement of $\Delta$, i.e. $X^2 - \Delta$) is open. This would mean that, $\forall z \in \Delta^C$, $\exists U_z$ such that $U_z \subseteq \Delta^C$.

So we take some point $z := (x,y) \in \Delta^C$. Then we know $x \ne y$ (it is in the complement of the set where $x=y$).

By assumption, $\{x\} = \bigcap \cal X$ and $\{y\} = \bigcap \cal Y$ (where $\cal Y$ is defined analogously to $\cal X$ in the obvious way). Then from the assumption and that $x,y$ are distinct, we know $x$ is in no closure of a neighborhood of $y$, and vice versa. (Symbolically, $\forall U_x$, $y \not \in \overline{U_x}$, and $\forall U_y$, $x \not \in \overline{U_y}$.)

I'm not really sure where to go from here, however. My feeling is that we need to take the product of two open sets (which itself is obviously open), and then show that it is a subset of $\Delta^C$ (or that the product has an empty intersection with $\Delta$, either way). The question is, what open sets? Presumably they would be neighborhoods of $x$ and $y$, and thus the product will be a neighborhood of $z = (x,y)$. But I'm not sure how to ensure that said product does not intersect $\Delta$. (Even if from an intuitive standpoint it does seem to be possible to construct such a product.) Besides, I'm not fully sure how the assumption would play into that.

Can anyone point me in the right direction as to what I should do?

Answer 1

Assume (4), and let $\langle x,y\rangle\in(X\times X)\setminus\Delta$. Then $x\ne y$, so $y\notin\bigcap\mathcal{X}$, and therefore there must be an open nbhd $U_x$ of $x$ such that $y\notin\operatorname{cl}U_x$. Let $V=X\setminus\operatorname{cl}U_x$; then $V$ is an open nbhd of $y$, so $U_x\times V$ is an open nbhd of $\langle x,y\rangle$. Clearly $U_x\cap V=\varnothing$, so $(U_x\times V)\cap\Delta=\varnothing$, $\langle x,y\rangle\notin\operatorname{cl}\Delta$, and $\Delta$ is closed.