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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Literature-sourced — the exact statement appears in a cited source; only wording and notation differ.
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The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset

Statement

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

If $(X,d_X)$ is compact (Open cover, subcover, compact metric space, and compact subset of a metric space), the image $f[X] \subseteq Y$ is a compact subset of $Y$ .
More generally, if $K \subseteq X$ is a compact subset of $X$ , then $f[K]$ is a compact subset of $Y$ .

No choice principle is used.

Facts & Assumptions

Given: Metric spaces $(X,d_X)$ and $(Y,d_Y)$ and a continuous $f : X \to Y$ ; images and preimages are written $f[\,\cdot\,]$ and $f^{-1}[\,\cdot\,]$ (Injection, surjection, bijection).

[L1]

A subset $A$ of a metric space is compact exactly when for every family $(V_i)_{i \in I}$ of open subsets of the ambient space with $A \subseteq \bigcup_{i \in I} V_i$ there are $n \in \mathbb{N}$ and $i_0, \dots, i_n \in I$ with $A \subseteq V_{i_0} \cup \dots \cup V_{i_n}$ , or else $A = \emptyset$ ; and a space is a compact subset of itself exactly when it is a compact metric space (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, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L2]

$f$ is continuous exactly when $f^{-1}[V]$ is open in $X$ for every open $V \subseteq Y$ (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, 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).

[L3]

The restriction of $f$ to a metric subspace $(K,d_K)$ of $X$ is continuous as a map $(K,d_K) \to (Y,d_Y)$ , since the $\varepsilon$ - $\delta$ condition at a point of $K$ is the condition for $f$ at that point read for the points of $K$ only, and $d_K$ is the restriction of $d_X$ (Isometry, isometric embedding, and the subspace metric on a subset, Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form).

Proof

technique · direct

1.1

Assume $(X,d_X)$ compact and let $(V_i)_{i \in I}$ be a family of open subsets of $Y$ with $f[X] \subseteq \bigcup_{i \in I} V_i$ .

2.1

Each $f^{-1}[V_i]$ is open in $X$ , and $X = \bigcup_{i \in I} f^{-1}[V_i]$ , because every $x \in X$ has $f(x) \in f[X]$ and so $f(x) \in V_i$ for some $i$ .

L2step 1.1

3.1

If $X = \emptyset$ then $f[X] = \emptyset$ and there is nothing to prove; otherwise compactness of $X$ , read against the indexed family of step 2.1, gives $n \in \mathbb{N}$ and $i_0, \dots, i_n \in I$ with $X = f^{-1}[V_{i_0}] \cup \dots \cup f^{-1}[V_{i_n}]$ .

L1step 2.1

4.1

Every $y \in f[X]$ is $f(x)$ for some $x \in X$ , and that $x$ lies in some $f^{-1}[V_{i_j}]$ , so $y \in V_{i_j}$ ; hence $f[X] \subseteq V_{i_0} \cup \dots \cup V_{i_n}$ and $f[X]$ is a compact subset of $Y$ : claim 1.

L1step 3.1

5.1

For claim 2, let $K \subseteq X$ be a compact subset, so that $(K,d_K)$ is a compact metric space; the restriction of $f$ to $K$ is continuous, and its image is $f[K]$ , so claim 1 applied to that restriction gives that $f[K]$ is a compact subset of $Y$ .

L1L3step 4.1∎

Remarks

Compactness travels forwards, not backwards. The preimage of a compact set under a continuous map need not be compact: a constant map from an unbounded space has a one-point image. What claim 1 uses is that preimages of open sets are open, which is the content of continuity, together with the fact that a finite subcover upstairs projects to a finite subcover downstairs.

Consequences on this page. Claim 1 with $Y = \mathbb{R}$ gives the extreme value theorem (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value), and claim 2 is what makes the inverse of a continuous bijection from a compact space continuous (A continuous bijection from a compact metric space onto a metric space carries open sets to open sets, so its inverse is continuous).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 75 results over 16 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)