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FALSE: every topology is induced by some metric

Statement

False claim: every topological space $(X,\mathcal{T})$ is metrizable, that is, for every topology there is a metric on $X$ whose metric topology is $\mathcal{T}$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

The claim fails for the smallest interesting reason available: every metric space separates distinct points by disjoint open sets (Distinct points of a metric space have disjoint balls around them), and the indiscrete topology on a set with two points has no two disjoint nonempty open sets at all (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

Facts & Assumptions

Given: A two-point set $X = \{a,b\}$ with $a \ne b$ , carrying the indiscrete topology $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ .

[A1]

The indiscrete topology on $X$ has exactly the two open sets $\varnothing$ and $X$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[A2]

$(X,\mathcal{T})$ is metrizable when some metric $d$ on $X$ has $\mathcal{T}_d = \mathcal{T}$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L1]

In a metric space, distinct points $p \ne q$ admit disjoint open sets $B(p,r)$ and $B(q,r)$ containing $p$ and $q$ respectively, with $r = d(p,q)/2 > 0$ (Distinct points of a metric space have disjoint balls around them).

[L2]

Open sets of a metric topology are exactly the members of $\mathcal{T}_d$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and a topology contains $\varnothing$ and $X$ (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

Refutation

technique · direct

1.1

Suppose $d$ were a metric on $X$ with $\mathcal{T}_d = \mathcal{T}_{\mathrm{ind}}$ .

assume-hyp

1.2

Since $a \ne b$ , [L1] supplies open sets $U \ni a$ and $V \ni b$ of $(X,d)$ with $U \cap V = \varnothing$ .

givenL1

2.1

By the supposition of step 1.1 the sets $U$ and $V$ lie in $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ ; and $a \in U$ , $b \in V$ make both nonempty, so $U = V = X$ by [A1].

step 1.1step 1.2A1L2

3.1

Then $U \cap V = X \ne \varnothing$ , since $a \in X$ , contradicting the disjointness of step 1.2; so no such metric $d$ exists and $(X, \mathcal{T}_{\mathrm{ind}})$ is not metrizable.

step 1.2step 2.1A2∎

Remarks

A second reading of the same failure, in terms of limits. In the indiscrete topology on $X$ every sequence converges to every point (Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure), whereas in a metric space a sequence has at most one limit (A sequence in a metric space has at most one limit). A constant sequence therefore converges to two distinct points here and could not do so under any metric. This is the same obstruction as the one used above, since uniqueness of limits is a consequence of the separation of distinct points by disjoint balls.
Separation is one obstruction and countability is another. Every metrizable space is Hausdorff, which is what the refutation above uses, and every metrizable space is also first countable (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, First countable space: a countable neighbourhood base at every point). Assuming the Axiom of Countable Choice, the cocountable topology on $\mathbb{R}$ fails the second: were it first countable, Assuming Countable Choice, in a first countable space sequential closure equals closure and sequential continuity at a point equals continuity there would make the identity onto the usual topology continuous, which it is not (FALSE: a sequentially continuous map between topological spaces is continuous). Under that assumption it is therefore not metrizable by this second obstruction as well. Neither obstruction is a characterisation: metrization theorems need separation and countability axioms beyond what is available at this point in the reading order.
What is being refuted is an existence claim, so the refutation must rule out every metric, which is why it argues from a property that all metric topologies share rather than by inspecting candidate metrics. The witness is worked again on the companion page (The indiscrete topology on a two-point set is induced by no metric ↗).

Depends on

Used by

The indiscrete topology on a two-point set is induced by no metric Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 95 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Metrizable space (Wikipedia) (standard reference, not scraped)
Hausdorff space (Wikipedia) (standard reference, not scraped)
Trivial topology (Wikipedia) (standard reference, not scraped)