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
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