Hausdorff via the Diagonal

Let $(X, \mathcal{T})$ and $(Y, \mathcal{T}_Y)$ be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Throughout, $X \times Y$ is the binary product $\prod_{i<2} X_i$ with $X_0 = X$ and $X_1 = Y$ (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), carrying the product topology; a point of it is a function $z$ on the von Neumann natural $2 = \{0,1\}$, written $(z_0, z_1)$, and $\pi_0, \pi_1$ are the two projections.

The basis used throughout. For the index set $2$ the product basis and the box basis coincide, since a box $\prod_{i<2} U_i$ has all but finitely many factors unrestricted for the trivial reason that it has only two (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). So

$\{\, U \times V : U \in \mathcal{T},\ V \in \mathcal{T}_Y \,\}$

is a basis for the product topology on $X \times Y$, and every statement below that tests a basic open set tests a box of two open sets.

The diagonal. The diagonal of $X$ is

$\Delta_X \;:=\; \{\, z \in X \times X : z_0 = z_1 \,\} \;=\; \{\, (x,x) : x \in X \,\} ,$

the second description being the first read through the definition of a point of the product as a function on $2$. It is a subset of $X \times X$ and is given the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) whenever it is regarded as a space.

The diagonal map. The diagonal map of $X$ is

$\delta_X : X \to X \times X, \qquad \delta_X(x) := (x,x) ,$

that is, the function sending $x$ to the constant function $2 \to X$ with value $x$. Its two components are $\pi_0 \circ \delta_X = \mathrm{id}_X$ and $\pi_1 \circ \delta_X = \mathrm{id}_X$, and by claim 2 of 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 it is the unique function $X \to X \times X$ with those two components. The same claim makes it continuous (Continuity of a map of topological spaces at a point and globally), the identity being continuous. Its image is $\Delta_X$, and it is injective, since $\delta_X(x) = \delta_X(x')$ forces $x = x'$ by reading the coordinate at $0$. Whether $\delta_X$ is an embedding onto $\Delta_X$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) is not asserted here; it is the content of the next item.

The pairing of two maps. For functions $f : Z \to X$ and $g : Z \to Y$ on a common domain, the pairing is

$\langle f, g \rangle : Z \to X \times Y, \qquad \langle f, g \rangle(z) := (f(z), g(z)) .$

By claim 2 of 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 it is the unique function $Z \to X \times Y$ with $\pi_0 \circ \langle f, g \rangle = f$ and $\pi_1 \circ \langle f, g \rangle = g$; no hypothesis on $f$ and $g$ is needed for the pairing to be defined, and continuity of the pairing is exactly continuity of both components, which is again that claim. In this notation

$\delta_X = \langle \mathrm{id}_X, \mathrm{id}_X \rangle ,$

so the diagonal map is a special case of the pairing and needs no separate treatment.

The preimage identity that every later proof uses. For $f, g : Z \to Y$,

$\langle f, g \rangle^{-1}[\Delta_Y] \;=\; \{\, z \in Z : f(z) = g(z) \,\} ,$

directly from the definitions above: $\langle f, g \rangle(z) \in \Delta_Y$ says that the function $(f(z), g(z))$ on $2$ takes the same value at $0$ and at $1$.

• The diagonal is a subset of a product, and the diagonal map is a function into it; they are different objects with the same name. The set $\Delta_X$ records which pairs are repetitions, and the map $\delta_X$ produces the repetitions. Both are needed: the closedness criterion of this page is about the set, and the transport of properties from $X$ to its copy inside the square is about the map.

• Nothing here depends on a choice principle. The product $X \times X$ is a binary product, and a point of it is exhibited by naming its two coordinates; the nonemptiness of an arbitrary product, which is where choice enters (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, claim 4), is never invoked for a binary product with a named point.

• Why the box description is recorded at the top. The criterion proved on this page tests basic open sets of $X \times X$, and for the binary product there is no gap between the box topology and the product topology to worry about (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). No infinite product is formed anywhere on this page, so the distinction never becomes live here.

Let $X$, $Y$ and $Z$ be topological spaces, with $X \times Y$ and $X \times X$ carrying the product topology and $\Delta_X$ the subspace topology (The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps, 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, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

1. The pairing is continuous exactly when both components are. For functions $f : Z \to X$ and $g : Z \to Y$, the pairing $\langle f, g \rangle$ is continuous if and only if $f$ and $g$ are continuous (Continuity of a map of topological spaces at a point and globally).
2. The diagonal map is an embedding. $\delta_X : X \to X \times X$ is injective and continuous, its image is $\Delta_X$, and the corestriction $\delta_X^{0} : X \to \Delta_X$, $\delta_X^{0}(x) = (x,x)$, is a homeomorphism. So $\delta_X$ is an embedding (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) and $X \cong \Delta_X$.
3. The inverse of $\delta_X^{0}$ is the restriction of the projection $\pi_0$ to $\Delta_X$, and this restriction agrees with the restriction of $\pi_1$.

Claim 2 is what licenses reading a property of $\Delta_X$ as a property of $X$: being a topological property is exactly invariance under homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

• The projection restricted to the diagonal is the inverse, and that is why no openness argument is needed. A continuous bijection is in general not a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological); here the candidate inverse is available for free as a restriction of a projection, so the homeomorphism is exhibited rather than deduced.

• Nothing in claim 2 uses a separation hypothesis. Every space, Hausdorff or not, sits inside its own square as the homeomorphic copy $\Delta_X$. What the Hausdorff condition decides is a different question, whether that copy is closed, and that is the next item.

• Claim 1 is the binary case of the characteristic property and is stated separately only because it is used constantly. For a product of two factors the general statement of 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 specialises to exactly the displayed equivalence, and the pairing notation is what makes the specialisation usable without re-indexing.

Let $(X, \mathcal{T})$ be a topological space and give $X \times X$ the product topology (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). Then $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) if and only if the diagonal $\Delta_X$ (The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps) is closed in $X \times X$:

$X \text{ Hausdorff} \iff \Delta_X = \overline{\Delta_X} \text{ in } X \times X .$

The condition on the right is a single closedness statement about one subset of one space, with no quantifier over pairs of points visible in it; that is what makes the criterion useful. In particular, the closed agreement-set result below is obtained by pulling $\Delta_X$ back along a continuous pairing, and the graph result is a specialization of that argument.

• The criterion is about the product topology on a binary product, and there the box basis and the product basis are the same family (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), so the boxes tested in steps 2.1 and 2.2 are legitimately basic. No infinite product is formed anywhere in the argument, and the criterion says nothing about one.

• Neither direction spends a choice principle. The forward direction produces one box from one Hausdorff separation of one named pair, and the backward direction reads one box out of the closure characterisation; there is no family to select from in either.

• What the criterion does not say. It does not say that $\Delta_X$ is closed in $X \times X$ carrying some other topology, and it does not say that $\Delta_X$ is closed in $X$ — the latter is not even a statement, $\Delta_X$ being a subset of the square. The hypothesis that $X \times X$ carries the product topology is used at [A2] and cannot be dropped.

Let $Z$ be a topological space, let $Y$ be a Hausdorff space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f, g : Z \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Then the agreement set

$E(f,g) \;:=\; \{\, z \in Z : f(z) = g(z) \,\}$

is closed in $Z$.

No hypothesis is placed on $Z$: the separation hypothesis is on the codomain alone, and it is not decoration. Let $Y_0 = \{a,b\}$ with $a \ne b$ carry the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which is not Hausdorff. Every function $Z \to Y_0$ is continuous, the only preimages to check being those of $\varnothing$ and $Y_0$, namely $\varnothing$ and $Z$. So for any subset $S \subseteq Z$ the constant map $f_0 \equiv a$ and the map $g_0$ taking the value $a$ on $S$ and $b$ off $S$ are continuous with $E(f_0,g_0) = S$, closed or not.

• Why the diagonal criterion is the right tool here. The condition "$f(z) = g(z)$" is a condition on the pair of values, so it becomes a membership condition once the two maps are packaged into one map into the square; the criterion then converts the separation hypothesis on $Y$ into the closedness of the set that condition names. Nothing is proved twice: the whole content is A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology together with the preimage identity of The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps.

• Both hypotheses are used, and only these. Continuity of $f$ and $g$ enters only through [L1], and the Hausdorff condition only through [L2]. In particular no countability, compactness or separation hypothesis on $Z$ appears anywhere in the argument.

• The complement is what the statement is often used for. $Z \setminus E(f,g)$ is open, so if $f$ and $g$ differ at a point they differ throughout some open neighbourhood of it. Equivalently, $E(f,g)$ contains the closure of every subset of $Z$ on which $f$ and $g$ agree, which is the form in which a statement about a dense set is obtained from this one.

Let $Z$ be a topological space, let $D \subseteq Z$ be dense (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets), let $Y$ be Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f, g : Z \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally) with

$f(d) = g(d) \qquad \text{for every } d \in D .$

Then $f = g$.

So a continuous map into a Hausdorff space is determined by its restriction to any dense subset of its domain. Nothing is asserted about which functions on $D$ extend: the statement is about uniqueness of an extension, not existence.

• The Hausdorff hypothesis is spent exactly once, inside [L1], and the density hypothesis exactly once, at step 2.1. Neither is used anywhere else, and neither can be weakened to the other: a dense agreement set alone does not force equality without a separation hypothesis on the codomain, and a Hausdorff codomain alone plainly does not.

• Density is a hypothesis about $Z$, not about $Y$. In particular the statement is about one domain and one dense subset of it; it says nothing about restrictions to subsets that are merely large in some other sense, and there is no cardinality condition anywhere in it.

• The uniqueness/existence split matters. A continuous $f : D \to Y$ need not extend continuously to $Z$ at all. What this corollary rules out is two different extensions, and that is exactly what makes an extension, when it exists, worth naming.

Let $X$ be a Hausdorff space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), let $(x_k)$ be a sequence in $X$ and let $p, q \in X$ with $x_k \to p$ and $x_k \to q$ (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure). Then $p = q$.

So in a Hausdorff space a sequence has at most one limit, and the notation $\lim_k x_k$ that Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure withholds in a general space is legitimate there.

The converse is false. Uniqueness of sequential limits does not imply the Hausdorff condition: the cocountable topology on $\mathbb{R}$ has unique sequential limits and is not Hausdorff (FALSE: a space in which every sequence has at most one limit is Hausdorff). So this lemma is strictly weaker than the hypothesis it is proved from, and it is not a characterisation.

• Why this is not the diagonal criterion in disguise. The criterion of this page characterises the Hausdorff condition exactly; the sequential statement above does not, and the gap is recorded by FALSE: a space in which every sequence has at most one limit is Hausdorff. A sequence sees only countably many points, and a space may separate no pair of points by open sets while still admitting no non-trivial convergent sequence at all.

• No choice principle is used. The two indices $K_1$ and $K_2$ come from two named neighbourhoods, and step 4.1 compares two given naturals; nothing is selected from a family.

• The statement is about limits, not about cluster points. A sequence in a Hausdorff space may well have many cluster points (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), and nothing above bears on that.

Let $X$ be a compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) topological space. Then:

1. $X$ is regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly);
2. $X$ is normal (Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly);
3. $X$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), and hence $X$ is $T_3$ and $T_4$.

Following Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly and Normal spaces and $T_4$ spaces, with the source disagreement over whether normality includes $T_1$ stated explicitly, regular and normal name the separation conditions alone and the numerals $T_3$ and $T_4$ name their conjunctions with $T_1$; claim 3 is what supplies the $T_1$ half, and it is stated separately for that reason.

Nothing stronger is claimed. In particular it is not asserted here that a compact Hausdorff space is completely regular (Regular spaces and $T_3$ spaces, with the source disagreement over whether regularity includes $T_1$ stated explicitly distinguishes the two conditions), and no continuous real-valued function is produced anywhere below.

• The whole content is that "closed" and "compact" coincide here, in the direction that is needed. Regularity asks a point to be separated from a closed set and normality asks two closed sets to be separated; compactness of the ambient space converts each closed set into a compact one, and the separation of compact sets in a Hausdorff space is what 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 supplies. No new separation argument is run.

• Why compactness of $X$ is needed and not just of the sets separated. The hypothesis is used only through [L1], to know that an arbitrary closed subset of $X$ is compact. A Hausdorff space in which the sets to be separated happen to be compact is separated by [L2] alone and needs no hypothesis on the ambient space at all; what compactness of $X$ buys is that every closed set is such a set.

• The degenerate cases are not a gap. If $C$, $A$ or $B$ is empty the required open sets are named outright in [A1] and [A2], so the argument does not depend on any nonemptiness hidden in the compact-separation clauses.

Let $X$ be a topological space, let $Y$ be Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f : X \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Then the graph

$G_f \;:=\; \{\, z \in X \times Y : z_1 = f(z_0) \,\} \;=\; \{\, (x, f(x)) : x \in X \,\}$

is closed in $X \times Y$ with the product topology (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).

No hypothesis is placed on $X$. The Hausdorff hypothesis is on the codomain and the continuity hypothesis is on $f$; both are used, and the converse implication — that a closed graph forces continuity — needs a different hypothesis on the codomain and is treated separately.

• The graph is an agreement set, and that is the whole proof. Writing $G_f$ as the set where $f \circ \pi_0$ and $\pi_1$ agree turns a statement about a map into a statement about two maps out of one space, which is exactly the shape For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set $\{ z \in Z : f(z) = g(z) \}$ is closed in $Z$ handles. Equivalently $G_f = \langle f \circ \pi_0, \pi_1 \rangle^{-1}[\Delta_Y]$, the preimage of the diagonal (The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps).

• The Hausdorff hypothesis is not removable. Let $X$ be a one-point space and let $Y = \{a,b\}$ with $a \ne b$ carry the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Every function $X \to Y$ is continuous, the only preimages to check being those of $\varnothing$ and $Y$. The product $X \times Y$ has as its open boxes only $\varnothing$ and itself, so its only closed sets are $\varnothing$ and itself; and $G_f$ is a single point, hence neither. The argument above breaks at [L3] and nowhere else.

• Continuity of $f$ is used, and only through the composite. Step 2.1 is the only appearance of the hypothesis; everything else is a property of the product.

Let $X$ and $Y$ be topological spaces, let $f : X \to Y$ be a function, and give $X \times Y$ the product topology (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), writing

$G_f \;=\; \{\, z \in X \times Y : z_1 = f(z_0) \,\}$

for the graph of $f$. Then:

1. Closed graph implies continuity, over a compact codomain. If $Y$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right) and $G_f$ is closed in $X \times Y$, then $f$ is continuous (Continuity of a map of topological spaces at a point and globally). No separation hypothesis on $Y$ is used in this direction.
2. Continuity implies closed graph, over a Hausdorff codomain. If $Y$ 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 $f$ is continuous, then $G_f$ is closed in $X \times Y$.
3. The equivalence. If $Y$ is compact and Hausdorff then $f$ is continuous if and only if $G_f$ is closed in $X \times Y$.