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CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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The identity from the discrete topology on R\mathbb{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)\mathcal{P}(\mathbb{R}) be the discrete topology on R\mathbb{R} (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and TR\mathcal{T}_{\mathbb{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\mathbb{R} a metric space: d(x,y)=xyd(x,y) = |x-y| is a metric, its open balls are the intervals (xr,x+r)(x-r, x+r), and it is unbounded). The identity id:(R,P(R))(R,TR)\mathrm{id} : (\mathbb{R}, \mathcal{P}(\mathbb{R})) \longrightarrow (\mathbb{R}, \mathcal{T}_{\mathbb{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\mathbb{R} that both arise in practice.

Facts & Assumptions

Given: R\mathbb{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\mathbb{R} is P(R)\mathcal{P}(\mathbb{R}): every subset is open (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies).

[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).

Counterexample

technique · direct
1.1

id\mathrm{id} is a bijection of R\mathbb{R} onto R\mathbb{R}, being the identity function of the set R\mathbb{R}.

given
1.2

id\mathrm{id} is continuous: for any open VV of the target the preimage id1[V]=V\mathrm{id}^{-1}[V] = V is a subset of R\mathbb{R} and hence open in the discrete topology.

A1L1
1.3

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

A1
1.4

{0}\{0\} is not open in the usual topology: for any r>0r > 0 the ball (r,r)(-r,r) contains the point 1/n1/n for a natural n1n \ge 1 with 1/n<r1/n < r supplied by [L4], and 1/n>01/n > 0, so 1/n(r,r)1/n \in (-r,r) and 1/n01/n \ne 0; hence no ball around 00 lies inside {0}\{0\}.

L3L4
2.1

By steps 1.3 and 1.4 the image id[{0}]={0}\mathrm{id}[\{0\}] = \{0\} of an open set is not open, so id\mathrm{id} 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

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