Why the criterion is about the product topology, and the choice cost of the compact separation lemmas

The criterion is a statement about the product topology, and the proof uses one specific fact about it. A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology tests the closedness of $\Delta_X$ against basic open sets of $X \times X$, and the basic open sets it uses are the boxes $U \times V$ with $U$ and $V$ open in $X$. That those boxes really are a basis is a feature of a binary product: by The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space a basic product-open set is a box all but finitely many of whose factors are unrestricted, and a box with two factors satisfies that condition for the trivial reason that it has only two. So for $X \times X$ the box basis and the product basis are one family, and the criterion carries no ambiguity about which of the two topologies is meant. No product with an infinite index set is formed anywhere on this page, and nothing here is asserted about one.

What the criterion buys, in one sentence. The Hausdorff condition is a quantifier over pairs of points and pairs of open sets; the criterion converts it into the closedness of a single subset of a single space. Every consequence on this page is then obtained the same way: package two maps into one map into a square with The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps, and pull $\Delta$ back along it, using the characteristic property of the product (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice) to know the packaged map is continuous. That is why an agreement set, and a graph, and the equality of two maps on a dense set are corollaries of the criterion rather than independent arguments.

The separation of compact sets, and what the naive proof of it would cost. The separation clauses used on this page are those of In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones: in a Hausdorff space a point and a disjoint compact set have disjoint open neighbourhoods, and so do two disjoint compact sets. The argument everyone writes first is

for each $y \in K$ choose disjoint open $U_y \ni x$ and $V_y \ni y$,

and it selects one pair of open sets for each point of an arbitrary set $K$. That is an application of the Axiom of Choice (The Axiom of Choice, Choice function), and it is avoidable. Take instead the family $\mathcal{V}$ of all open $V$ for which there exists an open $U$ with $x \in U$ and $U \cap V = \varnothing$. This family is specified by a formula, so nothing is selected in forming it; it covers $K$, because $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $x \notin K$; compactness (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) cuts it down to finitely many members $V_0, \dots, V_{n-1}$; and only now is a $U_i$ chosen for each $i < n$ — finitely many choices, licensed by Every natural-number-indexed list of nonempty sets has a choice function on its family of values, which is a theorem of ZF. Then $U := \{\, t \in X : t \in U_i \text{ for every } i < n \,\}, \qquad V := \bigcup_{i<n} V_i$ are the required neighbourhoods: $U$ is open, being $X$ when $n = 0$ and a finite intersection of open sets otherwise, it contains $x$, and it misses each $V_i$ because it is contained in each $U_i$. The same manoeuvre — collect a formula-defined family, cut it down by compactness, choose only afterwards — is what the closed-graph criterion on this page does. Where a step of this page spends a choice principle the step names it, and it is never more than Every natural-number-indexed list of nonempty sets has a choice function on its family of values.

Why the sequential form is weaker, and how much weaker. Uniqueness of sequential limits follows from the Hausdorff condition and does not imply it, which is why the criterion above is stated for the diagonal and not for sequences: a sequence sees at most countably many points, whereas closedness of $\Delta_X$ is a condition at every point of the square at once.

Conventions. The separation vocabulary used here — regular and normal as conditions on sets alone, $T_3$ and $T_4$ as their conjunctions with $T_1$ — is the one fixed in Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order, and every statement on this page writes the $T_1$ hypothesis out where it is used rather than building it into an adjective.