The Hausdorff property versus closedness of the diagonal in the context of convergence spaces Given a topological space $X$, the following are equivalent:


*

*Given points $x$ and $y$, there exist neighborhoods $A$ and $B$ of $x$ and $y$ respectively satisfying $A \cap B = \emptyset$.

*Every proper filter converges to at most one point.

*The diagonal set $\{(x,x) : x \in X\}$ is a closed subset of $X \times X$.


We learn as undergraduates that a topological space with any, and hence all, of these properties is said to be Hausdorff.
Now suppose we're trying to generalize from topological spaces to convergence spaces. The standard definition of being Hausdorff in this context is (2). Presumably, (1) is just the wrong definition in this context, and probably not worth thinking about too much. But (3) still looks interesting enough; for example, it's equivalent to being able to think of equality as a function $X \times X \rightarrow \Omega$, where $\Omega$ is the Sierpinski space.

Question. Is there a relationship between conditions (2) and (3) at the level of convergence spaces?

 A: These statements are still equivalent. We can apply the same idea than with topological spaces. Convergence of (proper) filters in a product space is component-wise. So a filter converges to two different points if and only if its image by the diagonal map $X \to X \times X$ converges to a point outside of the diagonal (that is, the diagonal is not closed).
We can do the same things with nets instead.
A: Just thinking out loud here...
One possible analogue could be to try if 
$X$ is a Hausdorff convergence space iff 

For any filter $\mathcal{F}$ on $X \times X$ such that $\Delta \in \mathcal{F}$, we can conclude from $\mathcal{F} \to (x,y)$ (in the product convergence) that $x = y$. This condition in fact says that the convergence-closure of $\Delta$ is $\Delta$ in the product convergence.

I don't know if this actually holds (I cannot show it, but I'm not very familiar with convergence spaces etc. beyond some definitions.) But is has the advantage it stays inside the category of convergence spaces. 
One direction I can show: suppose that $X$ is Hausdorff as a convergence.
Then $\Delta \subset X \times X$ is convergence-closed in the product convergence on $X \times X$.  Let $\mathcal{F}$ be a filter on $X \times X$ that contains $\Delta$ and such that $\mathcal{F} \to (x,y)$ in the product convergence. Let the projections be $\pi_1$ and $\pi_2$. Claim:
$$\pi_1[\mathcal{F}] = \pi_2[\mathcal{F}]$$
If $A \subseteq X$: $A \in \pi_1[F]$ iff $A \times X \in \mathcal{F}$ so $(A \times X) \cap \Delta \in \mathcal{F}$ and as $(A \times X) \cap \Delta = (X \times A) \cap \Delta  (=\{(x,x) \in X \times X: x \in A\})$ we have $(X \times A) \cap \Delta \in \mathcal{F}$ which implies $X\cap A \in \mathcal{F}$ so $A \in \pi_2[\mathcal{F}]$; the other inclusion goes the same way.
Let $\mathcal{G} = \pi_1[\mathcal{F}] = \pi_2[\mathcal{F}]$. Then the definition of product convergence says that $\mathcal{G} \to x$ (from the first projection) and also $\mathcal{G} \to y$ (from the second). As $X$ is Hausdorff as a convergence this can only happen if $x=y$ so that $(x,y) \in \Delta$. This shows that $\Delta$ is indeed convergence-closed in $X \times X$.
But we can also involve a topology:
For a convergence space we have an associated topology 
$$\mathcal{T} = \{O \subseteq X: \forall x \in O: \text{for all filters }\mathcal{F}: \text {if }\mathcal{F} \to x \text{ then } O \in \mathcal{F}\}$$
Then consider $Y = (X,\mathcal{T})^2$ in the product topology. Then $(X,\mathcal{T})$ is Hausdorff iff $\Delta$ is closed in $Y$.
We'd have something if $X$ is Hausdorff as a convergence space iff $(X,\mathcal{T})$ is Hausdorff. But the problem is (as I read online) that the convergence induced by $\mathcal{T}$ need not be the original convergence. It's only clear that $\mathcal{F} \to x$ implies $\mathcal{F} \rightarrow_{\mathcal{T}} x$, basically by definition. The reverse need not hold, I believe, and when the convergence we start with was actually induced from a topology, we do get this topology back in this way.
The Hausdorff equivalence probably doesn't really hold. 
We can show: 

if $(X,\mathcal{T})$ is Hausdorff then $X$ as a convergence space is Hausdorff.

This has basically the same proof as in topology: Assume $(X,\mathcal{T})$ is Hausdorff and suppose $\mathcal{F} \to x$ and $\mathcal{F} \to y$, $x \neq y$ in the convergence space. Then $\exists U_x \in \mathcal{T}: x \in U_x $ and $\exists U_y \in \mathcal{T}: y \in U_y$ such that $U_x \cap U_y = \emptyset$.
But the definition of $\mathcal{T}$ then says $U_x, U_y \in \mathcal{F}$ and we have a contradiction as their intersection is empty, which cannot happen in a filter.
The other direction is probably false. 
