TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A continuous bijection from a compact metric space onto a metric space carries open sets to open sets, so its inverse is continuous

Statement

Let $(X,d_X)$ be a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space), let $(Y,d_Y)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be a continuous bijection (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Injection, surjection, bijection). Then:

$f[U]$ is open in $Y$ for every $U$ open in $X$ (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 inverse function $f^{-1} : Y \to X$ is continuous.

The words used are deliberately those of open sets and of the inverse map: a single name for a continuous bijection with continuous inverse is not available at this point in the reading order. No choice principle is used.

Facts & Assumptions

Given: A compact metric space $(X,d_X)$ , a metric space $(Y,d_Y)$ and a continuous bijection $f : X \to Y$ .

[L1]

A closed subset of a compact metric space is a compact subset of it (A closed subset of a compact metric space is compact, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

[L2]

The image of a compact subset under a continuous map is a compact subset of the codomain (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

[L3]

A compact subset of a metric space is closed (A compact subset of a metric space is closed and bounded).

[L4]

A set is closed exactly when its complement is open (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).

[L5]

A map is continuous exactly when the preimage of every open set is open (For a map of metric spaces the following agree: $\varepsilon$ - $\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$ , Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form).

[L6]

For a bijection $f : X \to Y$ and $U \subseteq X$ : $f[X \setminus U] = Y \setminus f[U]$ , and for the inverse function $g = f^{-1}$ one has $g^{-1}[U] = f[U]$ (Injection, surjection, bijection).

Proof

technique · direct

1.1

Let $U \subseteq X$ be open; then $X \setminus U$ is closed in $X$ .

2.1

Being a closed subset of the compact space $X$ , the set $X \setminus U$ is a compact subset of $X$ .

L1step 1.1

3.1

Hence $f[X \setminus U]$ is a compact subset of $Y$ , and therefore closed in $Y$ .

L2L3step 2.1

4.1

Since $f$ is a bijection, $f[X\setminus U] = Y \setminus f[U]$ , so $f[U] = Y \setminus f[X \setminus U]$ is open in $Y$ : claim 1.

L4L6step 3.1

5.1

Write $g := f^{-1} : Y \to X$ , a function because $f$ is a bijection; for every open $U \subseteq X$ the preimage $g^{-1}[U]$ equals $f[U]$ , which is open by claim 1, so $g$ is continuous: claim 2.

L5L6step 4.1∎

Remarks

Compactness of the domain is essential. Without it a continuous bijection can have a discontinuous inverse, and no part of the argument survives, compactness being consumed at steps 2.1 and 3.1 alike. What the theorem says is that on a compact domain no such failure occurs, and the reason is entirely the open map property established at step 4.1.

Hausdorffness of the codomain is used silently and is automatic here. What step 3.1 needs is that a compact subset of $Y$ be closed, which is A compact subset of a metric space is closed and bounded and rests on the separation of distinct points by disjoint balls. Every metric space has that property, so no hypothesis on $(Y,d_Y)$ beyond being a metric space is required.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 17 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

Compact space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §26 (standard reference, not scraped)