Hausdorff via the Diagonal: Examples and Counterexamples

Give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and let $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ carry 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, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space). Then 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)

$\Delta_{\mathbb{R}} \;=\; \{\, (t,t) : t \in \mathbb{R} \,\}$

is closed in $\mathbb{R}^2$, and the box that separates a point $(a,b) \notin \Delta_{\mathbb{R}}$ from it may be written down:

$(a - r, a + r) \times (b - r, b + r), \qquad r := \tfrac{1}{2}|a - b| > 0 .$

Nothing here appeals to the general criterion; the computation is carried out against the product basis directly. It agrees with A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology, $\mathbb{R}$ being Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and the point of writing it out is to show what the criterion's abstract box is in this case: the two open intervals of half the distance between the coordinates.

• The radius $|a-b|/2$ is exactly the Hausdorff separation of $a$ and $b$ in $\mathbb{R}$, and that is not a coincidence. The proof of A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology builds its box out of a pair of disjoint open sets separating the two coordinates; here that pair is $(a-r, a+r)$ and $(b-r, b+r)$, the two balls of radius half the distance which the usual metric supplies.

• The product topology on $\mathbb{R}^2$ is the topology of the usual metrics on it, so the computation above may be read equally as a statement about boxes or about balls (For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space); the box form is used because it is what the criterion tests.

Standard topologies are as in The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, diagonals as in The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ of two maps, and every square carries 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).

1. Cofinite, on an infinite set. Let $X$ be infinite (Finite, countably infinite, countable, uncountable) and give it the cofinite topology $\mathcal{T}_{\mathrm{cof}}$. Then $(X, \mathcal{T}_{\mathrm{cof}})$ is $T_1$ ($T_0$ (Kolmogorov) and $T_1$ (Frechet) spaces), no two nonempty open sets are disjoint, $\overline{\Delta_X} \;=\; X \times X \;\ne\; \Delta_X ,$ so $\Delta_X$ is not closed and the space is not Hausdorff.
2. Cocountable, on $\mathbb{R}$. Give $\mathbb{R}$ the cocountable topology $\mathcal{T}_{\mathrm{coc}}$. The same three conclusions hold: $(\mathbb{R}, \mathcal{T}_{\mathrm{coc}})$ is $T_1$, no two nonempty open sets are disjoint, and $\overline{\Delta_{\mathbb{R}}} = \mathbb{R} \times \mathbb{R} \ne \Delta_{\mathbb{R}}$.
3. "Infinite" is the wrong hypothesis for the cocountable half. If $Z$ is countably infinite then $\mathcal{T}_{\mathrm{coc}}$ on $Z$ is the discrete topology, which is Hausdorff and whose diagonal is therefore closed. So clause 2 must be asserted of a set large enough that a cocountable set is a genuine restriction, and $\mathbb{R}$ is such a set; an arbitrary infinite set is not.

In every case the verdict on the diagonal matches A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology, as it must.

• Why clause 3 is stated rather than left implicit. The cofinite and the cocountable topologies behave alike only when the underlying set is large enough for the excluded sets to be a genuine restriction. On a countably infinite set "at most countable complement" excludes nothing, so the cocountable topology collapses to the discrete one and every conclusion of clause 2 reverses. Stating the two clauses with the same hypothesis would be a falsehood, and the falsehood is invisible unless the degenerate case is written out.

• The closure of the diagonal is as large as it can be. In both spaces of clauses 1 and 2 it is the entire square, so the diagonal is not merely non-closed: it is dense. That is the extreme opposite of the metric picture of The diagonal of $\mathbb{R}$ is closed in $\mathbb{R}^2$, computed from the product basis, where the diagonal is closed and its complement is open.

• $T_1$ is doing no work here. Both spaces satisfy $T_1$ and neither satisfies $T_2$, which is exactly the separation between the two axioms; the diagonal criterion detects the second and is blind to the first, since it is a statement about the square rather than about singletons.

Let $f : \mathbb{R} \to \mathbb{R}$ be continuous in the $\varepsilon$-$\delta$ sense of Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, give $\mathbb{R}$ its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) and give $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ 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, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$, and hence also of $d_1$ and $d_2$, so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space). Then the graph

$G_f \;=\; \{\, (x, f(x)) : x \in \mathbb{R} \,\}$

is closed in $\mathbb{R}^2$.

Every polynomial function is such an $f$, and so is every function built from continuous ones by the operations that preserve continuity; no further hypothesis on $f$ is needed, and in particular $f$ need not be bounded, monotone, or differentiable.

• What the $\varepsilon$-$\delta$ hypothesis becomes. The dictionary of Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$, continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace and Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not is what lets a hypothesis stated with $\varepsilon$ and $\delta$ be fed to a theorem stated about topological spaces; there is one notion of continuity for a real function here, not two, and step 1.2 is where that is used.

• The converse fails. A discontinuous $\mathbb{R} \to \mathbb{R}$ may still have closed graph — the function equal to $1/x$ off $0$ does (FALSE: every function between topological spaces whose graph is closed in the product is continuous) — so "closed graph" is strictly weaker than "continuous" for real functions. What restores the equivalence is a compact Hausdorff codomain (A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent).

• Nothing about $\mathbb{R}$ as a domain is used. The domain enters the argument only as an arbitrary topological space; the same proof gives a closed graph for a continuous map from any space into $\mathbb{R}$.

False claim: if $X$ is a topological space, $Y$ 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 $f : X \to Y$ has graph 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), then $f$ is continuous (Continuity of a map of topological spaces at a point and globally).

This is the sharpening of FALSE: every function between topological spaces whose graph is closed in the product is continuous that adds to the codomain the hypothesis under which the other half of the closed-graph criterion holds. It is still false, and the same witness refutes it:

$f : \mathbb{R} \to \mathbb{R}, \qquad f(x) = \frac{1}{x} \ \ (x \ne 0), \qquad f(0) = 0 ,$

with $\mathbb{R}$ carrying its usual topology, which is metrizable and hence Hausdorff. Its graph is closed in $\mathbb{R}^2$, it is continuous at every $c \ne 0$, and it is not continuous at $0$; so its set of discontinuities is exactly $\{0\}$.

What the criterion of A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent asks of the codomain in the direction "closed graph implies continuous" is compactness, and the Hausdorff condition contributes nothing there.

• Adding a separation hypothesis to the codomain cannot repair the claim, and this is why. In A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent the Hausdorff condition is what makes a continuous map have closed graph, and compactness is what makes a closed-graph map continuous. The two hypotheses belong to opposite directions, and the witness above has the first without the second.

• The failure is a single point, and it is not removable by redefining $f$ there. No value at $0$ makes $f$ continuous, because $|f(x)|$ exceeds every bound as $x$ approaches $0$; and no value at $0$ destroys the closedness of the graph. The example is therefore not a matter of a badly chosen value: it is the behaviour of the reciprocal near $0$, and a compact codomain is exactly what would forbid that behaviour.

• Where this sits relative to the functional-analytic closed graph theorem. That theorem replaces compactness of the codomain by completeness of both spaces and linearity of the map, and neither hypothesis is available or claimed here; the witness above is not linear, and nothing on this page bears on the functional-analytic statement.

Let $f, g : \mathbb{R} \to \mathbb{R}$ be continuous (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point) and suppose

$f(q) = g(q) \qquad \text{for every } q \in \mathbb{Q}_{\mathbb{R}} ,$

where $\mathbb{Q}_{\mathbb{R}}$ is the set of rationals inside $\mathbb{R}$ (The rationals embed densely in the reals). Then $f = g$.

So a continuous real function is determined by its values at the rationals, and two continuous functions that are visibly different must already differ at some rational. Nothing is claimed about which functions on $\mathbb{Q}_{\mathbb{R}}$ extend continuously to $\mathbb{R}$; the statement is about uniqueness of the extension only.

• Why the density is established from the order rather than quoted. What Two continuous maps into a Hausdorff space that agree on a dense subset are equal needs is density in the sense of Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets, a condition on the open sets of $\mathbb{R}$; step 1.1 derives exactly that condition from the statement that a rational lies strictly between any two reals (The rationals embed densely in the reals), which is where the order of $\mathbb{R}$ enters and the only place it does.

• Both hypotheses on the codomain and on the maps are needed. Dropping continuity of one map leaves the conclusion false for the obvious reason, and dropping the Hausdorff condition on the codomain leaves it false for a less obvious one: FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain exhibits two continuous maps on $\mathbb{R}$ agreeing at every rational, differing at every irrational, and taking values in a two-point space that is not Hausdorff.

• The rationals enter at one step only. They are used in step 1.1 and nowhere else; the argument as written proves the statement for an arbitrary dense subset of $\mathbb{R}$ once density in the sense of Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets is in hand, and step 1.1 is simply where that is established for $\mathbb{Q}_{\mathbb{R}}$.

False claim: for continuous maps $f, g : Z \to Y$ between topological spaces the agreement set $E(f,g) = \{\, z \in Z : f(z) = g(z) \,\}$ is closed in $Z$, with no hypothesis on the codomain $Y$.

The witness is the pair of maps of FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain: take $Z = \mathbb{R}$ with its usual topology, $Y_0 = \{a,b\}$ with $a \ne b$ and the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), $f \equiv a$, and $g$ equal to $a$ at every rational and to $b$ at every irrational. Both are continuous, and

$E(f,g) \;=\; \mathbb{Q}_{\mathbb{R}} ,$

which is dense in $\mathbb{R}$ (Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets) and is not all of $\mathbb{R}$, hence is not closed.

So the hypothesis dropped is the Hausdorff condition on the codomain, which is what For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set $\{ z \in Z : f(z) = g(z) \}$ is closed in $Z$ assumes; and the failure is the worst possible one, the agreement set being dense rather than merely non-closed.

• This is the sharp form of the failure recorded by FALSE: two continuous maps that agree on a dense subset of their common domain are equal, with no hypothesis on the codomain. That item refutes the equality of two maps agreeing on a dense set; the present one locates the reason, namely that without a separation hypothesis on the codomain the agreement set need not be closed, so a dense agreement set need not be all of the domain. Both failures come from the same pair of maps.

• Any subset of $\mathbb{R}$ can be an agreement set here. Replacing $\mathbb{Q}_{\mathbb{R}}$ by an arbitrary $S \subseteq \mathbb{R}$ in the definition of $g$ gives $E(f,g) = S$ by the computation of step 2.2, and $g$ is still continuous by step 2.1. So into a two-point indiscrete codomain the agreement set carries no topological information whatsoever, which is the strongest way the claim can fail.

• Why a two-point indiscrete codomain is the natural witness. A space with at most one point is Hausdorff vacuously (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not), so a witness needs at least two points; and the indiscrete topology is the coarsest topology on a two-point set, which is what makes every map into it continuous and so removes any need to verify continuity. Other non-Hausdorff topologies on two points exist — Sierpinski space is one (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) — and would require that verification instead.

Let $(X, \mathcal{T})$ 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) whose underlying set is finite (Finite, countably infinite, countable, uncountable). Then:

1. $\mathcal{T}$ is the discrete topology on $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies): every subset of $X$ is open.
2. $X \times X$ 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) is discrete as well, so every subset of it — the diagonal $\Delta_X$ included — is both open and closed.

Clause 2 makes the diagonal criterion (A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology) true here for a reason that has nothing to do with the diagonal: in a discrete square every subset is closed. The example is worth recording precisely because it is the degenerate case, where the criterion carries no information.

• The finiteness is used only through "every subset is finite". Nothing about cardinality beyond that enters, and the argument gives, for an arbitrary $T_1$ space, that every finite subset is closed — which is the content of clause (c) of A space is $T_1$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology and is the real reason a finite $T_1$ space is discrete.

• The Hausdorff hypothesis may be weakened to $T_1$. Step 1.1 is the only place it is used, and it is used only to obtain $T_1$; so a finite $T_1$ space is already discrete, and a finite Hausdorff space is discrete because Hausdorff implies $T_1$. Neither hypothesis can be dropped altogether: the indiscrete topology on a two-point set is finite and not discrete (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

• Why the diagonal is uninformative here. In a discrete square every subset is closed, so the closedness of $\Delta_X$ is not evidence of anything about $X$; the criterion is a genuine test only where the square has proper nonempty non-closed subsets to be distinguished from.

Let $X$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $F \subseteq X$ be finite (Finite, countably infinite, countable, uncountable), with 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). Then:

1. $F$ is compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right), whatever $X$ is and whatever topology it carries.
2. Consequently, if $X$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) then a point $x \in X \setminus F$ and the set $F$ have disjoint open neighbourhoods, and two disjoint finite subsets of $X$ have disjoint open neighbourhoods (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 particular $F$ is closed in $X$.

Clause 1 spends a choice principle, and exactly one: finite choice (Every natural-number-indexed list of nonempty sets has a choice function on its family of values), which is a theorem of ZF. The naive phrasing of the same argument — "for each $y \in F$ pick a member of the cover containing it" — is a selection over the index set of $F$, and because that index set is a natural number the selection is licensed outright.

• Clause 2 recovers the $T_1$ behaviour of a Hausdorff space by a different route. That every finite subset of a Hausdorff space is closed is usually read off from the separation axioms; here it arrives as a special case of a compactness statement, and the two readings agree, as they must.

• Where the finiteness of $F$ is used. Only in step 2.1, and only to make the selection a finite one. The same argument with an infinite $F$ would need a genuine choice principle and would in any case fail at step 3.1, an infinite index set producing no finite subcover.

• The example is the smallest non-trivial instance of the compact separation clauses. It needs no compactness hypothesis on $X$ and no cover argument beyond the one above, so it is the case in which the clauses 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 can be checked against intuition before being used on genuinely compact sets.