CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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The identity from the discrete topology on $R$ to the usual topology is a continuous bijection that is not a homeomorphism

Statement refuted

Refuted: that every continuous bijection of topological spaces is a homeomorphism (FALSE: every continuous bijection of topological spaces is a homeomorphism).

Witness. Let $P (R)$ be the discrete topology on $R$ (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and $T_{R}$ its usual topology (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $R$ 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). The identity $i d : (R, P (R)) ⟶ (R, T_{R})$ is a continuous bijection, is not an open map, and is therefore not a homeomorphism (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces).

The two-point witness inlined in the refutation of FALSE: every continuous bijection of topological spaces is a homeomorphism shows that the failure occurs in the smallest possible space; the present one shows that it occurs between two topologies on $R$ that both arise in practice.

Facts & Assumptions

Given: $R$ carrying the discrete topology as source and the usual topology as target, and the identity function between them.

[A1]

The discrete topology on $R$ is $P (R)$ : every subset is open (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[L1]

A map is continuous exactly when preimages of open sets are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b), Continuity of a map of topological spaces at a point and globally); an open map carries open sets to open sets (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[L2]

A continuous bijection is a homeomorphism if and only if it is an open map (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, claim 1).

[L3]

In the usual topology $B (x, r) = (x - r, x + r)$ , and $U$ is open exactly when every $x \in U$ has some $r > 0$ with $(x - r, x + r) \subseteq U$ (The absolute value makes $R$ 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, Open ball, closed ball and sphere in a metric space, 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, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / n < ε$ , and $1 / n > 0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

Counterexample

technique · direct

1.1

$i d$ is a bijection of $R$ onto $R$ , being the identity function of the set $R$ .

given

1.2

$i d$ is continuous: for any open $V$ of the target the preimage ${i d}^{- 1} [V] = V$ is a subset of $R$ and hence open in the discrete topology.

A1L1

1.3

${0}$ is open in the discrete topology by [A1].

1.4

${0}$ is not open in the usual topology: for any $r > 0$ the ball $(- r, r)$ contains the point $1 / n$ for a natural $n \geq 1$ with $1 / n < r$ supplied by [L4], and $1 / n > 0$ , so $1 / n \in (- r, r)$ and $1 / n \neq 0$ ; hence no ball around $0$ lies inside ${0}$ .

L3L4

2.1

By steps 1.3 and 1.4 the image $i d [{0}] = {0}$ of an open set is not open, so $i d$ is not an open map; with steps 1.1 and 1.2 it is a continuous bijection, so by [L2] it is not a homeomorphism, and equivalently its inverse is not continuous.

step 1.1step 1.2step 1.3step 1.4L1L2∎

Remarks

The two spaces are not homeomorphic at all, not merely not homeomorphic by this map. In the discrete topology every subset is clopen (The discrete and indiscrete topologies, their closures and interiors, and their continuous maps in each direction), while in the usual topology ${0}$ is closed and not open by step 1.4 above; "every subset is clopen" is a topological property (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), so no homeomorphism between them exists.
Both are metrizable, the discrete topology by the metric taking the value $1$ on distinct points and $0$ on equal ones, whose ball of radius $1$ about $x$ is ${x}$ , and the usual topology by $∣ x - y ∣$ (The absolute value makes $R$ 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). So metrizability of source and target is no help: the failure is about which topology, not about whether a metric exists.
The general pattern. Any two comparable and distinct topologies on one set give such a witness, the identity from the finer to the coarser (FALSE: every continuous bijection of topological spaces is a homeomorphism); the discrete topology is the finest of all (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), so it pairs with every non-discrete topology on $R$ .

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Sources

Homeomorphism (Wikipedia) (standard reference, not scraped)
Discrete space (Wikipedia) (standard reference, not scraped)

The identity from the discrete topology on RR to the usual topology is a continuous bijection that is not a homeomorphism