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Continuous Images of Compact Spaces

topologyUpdated 2026-09-17

Helpful firstCompactness ↗

Continuous maps carry compactness forward.

Theorem

If $X$ is compact and $f:X\to Y$ is continuous, then $f(X)$ is compact with the subspace topology.

Proof

Let $\{V_i\}$ be an open cover of $f(X)$ in the subspace topology. Their inverse images $f^{-1}(V_i)$ are open in $X$ and cover $X$ .

Compactness gives a finite subcover $f^{-1}(V_{i_1}),\ldots,f^{-1}(V_{i_n})$ . The corresponding $V_{i_1},\ldots,V_{i_n}$ cover $f(X)$ .

Remark

No Hausdorff assumption on $Y$ is needed for this statement.

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