# continuous image of a compact set is compact

###### Theorem 1.

The continuous^{} image of a compact set is also compact.

###### Proof.

Let $X$ and $Y$ be topological spaces^{}, $f:X\to Y$ be continuous, $A$ be a compact subset of $X$, $I$ be an indexing set, and ${\{{V}_{\alpha}\}}_{\alpha \in I}$ be an open cover of $f(A)$. Thus, $f(A)\subseteq {\displaystyle \bigcup _{\alpha \in I}}{V}_{\alpha}$. Therefore, $A\subseteq {f}^{-1}\left(f(A)\right)\subseteq {f}^{-1}\left({\displaystyle \bigcup _{\alpha \in I}}{V}_{\alpha}\right)={\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}({V}_{\alpha})$.

Since $f$ is continuous, each ${f}^{-1}({V}_{\alpha})$ is an open subset of $X$. Since $A\subseteq {\displaystyle \bigcup _{\alpha \in I}}{f}^{-1}({V}_{\alpha})$ and $A$ is compact, there exists $n\in \mathbb{N}$ with $A\subseteq {\displaystyle \bigcup _{j=1}^{n}}{f}^{-1}\left({V}_{{\alpha}_{j}}\right)$ for some ${\alpha}_{1},\mathrm{\dots},{\alpha}_{n}\in I$. Hence, $f(A)\subseteq f\left({\displaystyle \bigcup _{j=1}^{n}}{f}^{-1}({V}_{{\alpha}_{j}})\right)=f\left({f}^{-1}\left({\displaystyle \bigcup _{j=1}^{n}}{V}_{{\alpha}_{j}}\right)\right)\subseteq {\displaystyle \bigcup _{j=1}^{n}}{V}_{{\alpha}_{j}}$. It follows that $f(A)$ is compact. ∎

Title | continuous image of a compact set is compact |
---|---|

Canonical name | ContinuousImageOfACompactSetIsCompact |

Date of creation | 2013-03-22 15:53:14 |

Last modified on | 2013-03-22 15:53:14 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 16 |

Author | Wkbj79 (1863) |

Entry type | Theorem |

Classification | msc 54D30 |

Related topic | CompactnessIsPreservedUnderAContinuousMap |