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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?2012-07-26
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    Incidentally, your well-known result is usually proved using your later, true, result.2012-07-26
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    The continuous image of a compact set is compact.............2012-07-27

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