# Image of compacta under a continuous map

There is a well-known result in topology,

Any continuous bijection from a compact topological space to a Hausdorff space is a homeomorphism.

I was wondering whethet the following (slightly weaker) statement holds:

Let $K$ be a compact topological space and $X$ a topological space. Then $f(K)$ is compact in $X$ for any continuous map $f\colon K\to X$.

• Yes. Take an open cover of $f(K)$ and take preimages. A finite number of these cover $K$. Now what? Jul 26, 2012 at 23:52
• Incidentally, your well-known result is usually proved using your later, true, result. Jul 26, 2012 at 23:55
• The continuous image of a compact set is compact.............
– user38268
Jul 27, 2012 at 0:07

Take any be an open cover $\mathcal U$ of $f[K]$. Then, by continuity of $f$, the set $f^{-1}[U]$ is open in $K$ for any $U\in\mathcal U$. Thus, $\{f^{-1}[U]: U\in\mathcal U\}$ is an open cover of $K$.
By using that assumption that $K$ is compact, there is a finite sub-cover $\mathcal V\subset\mathcal U$ such that $\{f^{-1}[V]: V\in\mathcal V\}$ is a cover of $K$. Then $\{V:V\in\mathcal V\}$ covers $f[K]$, so $f[K]$ is compact.
This is true. Suppose $\{U_i\}_{i\in I}$ is an open cover of $f(K)$. Then $\{f^{-1}(U_i)\}_{i\in I}$ is an open cover of $K$, so has a finite subcover, say $f^{-1}(U_1),\ldots,f^{-1}(U_n)$. Then $U_1,\ldots,U_n$ is a finite subcover of $\{U_i\}_{i\in I}$, thus $f(K)$ is compact.
It holds if we add : $X$ a Hausdorff space