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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 $f\colon K\to X$.

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Yes. Take an open cover of $f(K)$ and take preimages. A finite number of these cover $K$. Now what? – Dylan Moreland Jul 26 '12 at 23:52
Incidentally, your well-known result is usually proved using your later, true, result. – David Mitra Jul 26 '12 at 23:55
The continuous image of a compact set is compact............. – user38268 Jul 27 '12 at 0:07

The statement is true:

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.

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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.

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It holds if we add : $X$ a Hausdorff space

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1 ws find : Some branches of mathematics such as algebraic geometry, typically influenced by the French school of Bourbaki, use the term quasi-compact for the general notion, and reserve the term compact for topological spaces that are both Hausdorff and quasi-compact. A single compact set is sometimes referred to as a compactum; following the Latin second declension (neuter), the corresponding plural form is compacta – Mohamed Jul 27 '12 at 0:03

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