CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The indiscrete topology on a two-point set is induced by no metric

Statement refuted

Refuted: that every topology is induced by some metric (FALSE: every topology is induced by some metric).

Witness. Let $X = \{a,b\}$ with $a \ne b$ , carrying the indiscrete topology $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). No metric $d$ on $X$ satisfies $\mathcal{T}_d = \mathcal{T}_{\mathrm{ind}}$ (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), so $(X,\mathcal{T}_{\mathrm{ind}})$ is not metrizable.

Facts & Assumptions

Given: The set $X = \{a,b\}$ with $a \ne b$ and the topology $\mathcal{T}_{\mathrm{ind}} = \{\varnothing, X\}$ ; and a hypothetical metric $d$ on $X$ with $\mathcal{T}_d = \mathcal{T}_{\mathrm{ind}}$ .

[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, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[A2]

A space is metrizable when some metric on it has the given topology as its metric topology (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, Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

In any metric space, distinct points $p \ne q$ satisfy $B(p,r) \cap B(q,r) = \varnothing$ for $r := d(p,q)/2 > 0$ , and these two balls are open and contain $p$ and $q$ respectively (Distinct points of a metric space have disjoint balls around them, Open ball, closed ball and sphere in a metric space).

Counterexample

technique · direct

1.1

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

assume-hyp

1.2

Since $a \ne b$ , [L1] gives $r := d(a,b)/2 > 0$ and two disjoint sets $U := B(a,r)$ and $V := B(b,r)$ , open in $(X,d)$ , with $a \in U$ and $b \in V$ .

givenL1

2.1

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

step 1.1step 1.2A1

3.1

Then $U \cap V = X$ , which contains $a$ and is therefore nonempty, contradicting the disjointness of step 1.2. So no such metric exists, and $(X, \mathcal{T}_{\mathrm{ind}})$ is not metrizable, which refutes the claim.

step 1.2step 2.1A2∎

Remarks

A second route, through limits. In the indiscrete topology every sequence converges to every point (The sequential closure is contained in the closure, continuity implies sequential continuity, and sequential limits need not be unique, claim 3, 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). The constant sequence at $a$ therefore has two limits here and could have only one under any metric. This is the same obstruction, since uniqueness of metric limits is proved from the separation of [L1].
Two points are the minimum. On a one-point set the indiscrete topology is metrizable, by the unique metric $d(x,x) = 0$ ; the failure needs two distinct points, and it needs them only to have distinct distance.
The obstruction used here is separation, not size. An uncountable metrizable space exists ( $\mathbb{R}$ ), and a finite non-metrizable space exists (this one), so cardinality is irrelevant. Assuming the Axiom of Countable Choice, the other obstruction developed on these pages — failure of first countability — also rules out the cocountable topology on $\mathbb{R}$ (In the cocountable topology on $\mathbb{R}$ the closed sets are the countable sets and $\mathbb{R}$ , and a sequence converges iff it is eventually constant, First countable space: a countable neighbourhood base at every point).

Depends on

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Sources

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