# the continuous image of a compact space is compact

Consider $f:X\to Y$ a continuous^{} and surjective function and $X$ a compact set. We will prove that $Y$ is also a compact set.

Let $\{{V}_{a}\}$ be an open covering of $Y$. By the continuity of $f$ the pre-image by $f$ of any open subset (http://planetmath.org/OpenSubset) of $Y$ will also be an open subset of $X$. So we have an open covering $\{{U}_{a}\}$ of $X$ where ${U}_{a}={f}^{-1}({V}_{a})$.

To see this remember, since the continuity of $f$ implies that each ${U}_{a}$ is open, all we need to prove is that ${\bigcup}_{a}{U}_{a}\supset X$. Consider $x\in X$, we know that since $\{{V}_{a}\}$ is a covering of $Y$ that there exists $i$ such that $f(x)\in {V}_{i}$ but then by construction $x\in {U}_{i}$ and $\{{U}_{a}\}$ is indeed an open covering of $X$.

Since $X$ is compact we can consider a finite set^{} of indices $\{{a}_{i}\}$ such that $\{{U}_{{a}_{i}}\}$ is a finite open covering of $X$, but then $\{{V}_{{a}_{i}}\}$ will be a finite open covering of $Y$ and it will thus be a compact set.

To see that $\{{V}_{{a}_{i}}\}$ is a covering of $Y$ consider $y\in Y$. By the surjectivity of $f$ there must exist (at least) one $x\in X$ such that $f(x)=y$ and since $\{{U}_{{a}_{i}}\}$ is a finite covering of $X$, there exists $k$ such that $x\in {U}_{{a}_{k}}$. But then since $f({U}_{{a}_{k}})={V}_{{a}_{k}}$, we must have that $y\in {V}_{{a}_{k}}$ and $\{{V}_{{a}_{i}}\}$ is indeed a finite open covering of $Y$.

Title | the continuous image of a compact space is compact |
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Canonical name | TheContinuousImageOfACompactSpaceIsCompact |

Date of creation | 2013-03-22 15:52:48 |

Last modified on | 2013-03-22 15:52:48 |

Owner | cvalente (11260) |

Last modified by | cvalente (11260) |

Numerical id | 7 |

Author | cvalente (11260) |

Entry type | Proof |

Classification | msc 54D30 |

Related topic | CompactnessIsPreservedUnderAContinuousMap |