# 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 TheContinuousImageOfACompactSpaceIsCompact 2013-03-22 15:52:48 2013-03-22 15:52:48 cvalente (11260) cvalente (11260) 7 cvalente (11260) Proof msc 54D30 CompactnessIsPreservedUnderAContinuousMap