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There is a similar question here: Proving a certain implication in the equivalent formulations of Hausdorff spaces but it does not ask for proof of this exact equivalence that I am asking.

My book has the following equivalences for a Hausdorff topological space.

(1) $X$ is Hausdorff

(2) The diagonal relation set $Δ := \left\{ (x,x) | x \in X \right\}$ is a closed set in $X^2$.

(3) Limits of nets in $X$ are unique

I'm looking for a direct proof of (2) => (3).

For (3) => (2) I have the following:

To show Δ is closed, we show that Δ = $\overline{Δ}$.

$\subset$ is always true.

$\supset$: Let $(a, b) \in \overline{Δ}$. Then there is a net in Δ, {$x_j$} that converges to $(a,b)$. Now since {$x_j$} is a net in Δ, for each index $j$, $x_j = (z_j, z_j)$. Observe that in both coordinates, the net in $X$ is exactly the same. So, for the first coordinate, $z_j$ converges to $a$ and for the second coordinate, $z_j$ converges to $b$. Now, by (3), the limit is unique, therefore $a = b$, thus $(a,b) \in Δ$.

The book I use do (1) => (3) => (2) => (1) so I actually do have a roundabout proof for what I want. But is there a direct way?

Teo LC
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  • Which book is it? – R. H. Vellstra Mar 31 '24 at 02:10
  • @pyridoxal_trigeminus it is a book written by my professor for our General Topology course. It is based off of Munkres but does not follow the exact order. – Teo LC Mar 31 '24 at 02:17
  • Case in point, we do not deal with countability axioms in the course. We do use the axiom of choice though when constructing counter examples. Connectedness and compactness come after separation axioms in the course. This differs from the order and depth in Munkres. – Teo LC Mar 31 '24 at 02:26
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    To show $(2)\implies(3)$ show the contrapositive. If $(x_{\lambda}){\lambda\in\Lambda}$ is a net in $X$ with at least two limits $a\neq b$, then look at $(x{\lambda},x_\lambda)$ to show that $(a,b),(b,a)\in\overline{\Delta}$ – Alessandro Codenotti Mar 31 '24 at 10:52

1 Answers1

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Let $y,z$ be limits of the net $x_\lambda$. Then it follows that $\langle y,z\rangle\in\operatorname{cl}\Delta$: given neighborhoods $U,V$ of $y,z$, we have $x_{\lambda}\in U$ for $\lambda\geq \lambda_x$ and $x_{\lambda}\in V$ for $\lambda\geq \lambda_y$. Then choose $\lambda'\geq\lambda_y,\lambda_z$ and we have $\langle x_{\lambda'},x_{\lambda'}\rangle\in U\times V$.

Since $\langle y,z\rangle\in\operatorname{cl}\Delta=\Delta$ by (2), we have $y=z$, thus (3).

Steven Clontz
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