---
tag: 0K9AAJ8G51
kind: statement
role: proposition
name: Continuous image of a compact set is compact
msc: [54D30]
uses: [0SY9ECQ19Q, 0P6PMJB4MA]
sources:
  - url: https://en.wikipedia.org/wiki/Compact_space
    locator: Functions and compact spaces
---

## Statement

Let $X$ and $Y$ be [[topological spaces|0CNQXVBHCX]], let $f \colon X \to Y$ be a [[continuous map|0SY9ECQ19Q]] and let $K \subseteq X$ be [[compact|00R0X1WGPX]].
Then the [[image|0Q3F20XB0A]] $f(K)$ is compact in $Y$.

## Remarks

Proof: let $(V_i)$ be open sets of $Y$ that cover $f(K)$. The sets $f^{-1}(V_i)$ are open and cover $K$. A finite subfamily covers $K$, and the matching $V_i$ cover $f(K)$.
A consequence is that a [[continuous real function on a compact set attains its bounds|0V4DET59NW]].
Compactness is a [[topological property|02SMVGC690]].

## Explanations

### L1 human

Continuous maps keep compactness.
For example, the image of $[0, 1]$ under a continuous map into the plane is a compact curve, so it is closed and bounded.