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I want to find theorems of the form "all continuous functions defined on a topological space X satisfy some properties if and only if $X$ is compact." Some restrictions on the space X may be acceptible, such as Hausdorff, metric space, etc. The codomain of the functions could be anything.

From the comments, I found that a metric space is compact if and only if every continuous real function on it is bounded. I would appreciate any other similar theorems.

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    Such a property is boundedness. – 2017-02-15
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    Boundedness isn't enough, all continuous functions $[0,\omega_1)\to\Bbb R$ are bounded but $[0,\omega_1)$ isn't compact – 2017-02-15
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    Boundedness of all real functions actually is exactly what is covered in this question: [Compact metric space characterization (continuous real functions)](http://math.stackexchange.com/q/1580657/214724) – 2017-02-15
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    If $X$ is compact Hausdorff and $Y$ is Hausdorff then any continuous $f:X\to Y$ is a closed mapping. If $X$ is Hausdorff but not compact does there exist a Hausdorff $Y$ and a continuous $f:X\to Y$ that is not a closed mapping? – 2017-02-15
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    @user254665 This seems like a variation (strengthening) of H-closedness. It might imply compactness, or it might not. Proofs? – 2017-02-15
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    @HennoBrandsma. I am not familiar with H-closedness. My Q is readily answered "Yes" if $X$ is regular as we can let $f=id_X$ where $X$ is dense in a compact space $Y.$ (So the image of $X$ is not closed in $Y.$) I dk he general answer. – 2017-02-18
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    @user254665 a regular space need not be a subspace of a compact Hausdorff space. So how would that work? – 2017-02-18
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    @I meant compact in the mot general sense, not necessarily Hausdorff. And in that sense I just noticed that my Q has an obvious answer: Let $\tau_X$ be the topology on $X.$ Let $p\not \in X.$ Let the topology on $Y=X\cup \{p\}$ be $\tau_X \cup \{Y\}.$ And $f=id_X.$ – 2017-02-18

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There is no such characterisation that I know of, but there are some known things:

A space $X$ is called pseudocompact iff every continuous $f:X \rightarrow \mathbb{R}$ is bounded.

For normal spaces ($T_1$ plus being able to separate disjoint closed sets) this is equivalent to countable compactness (every countable open cover has a finite subcover) and limit point compactness (every infinite subset has a limit point). For metric spaces all three are equivalent to compactness and sequential compactness as well (every sequence has a convergent subsequence). But not in general, as $\omega_1$, the first uncountable ordinal in the order topology or $\Sigma$-Products of uncountably copies of $[0,1]$ show.

These are normal, pseudocompact (and so countably compact etc.) but not compact.

Using mappings of a special kind : $X$ is compact iff for every space $Y$, the projection $\pi_Y: X \times Y \rightarrow Y$ is a closed map.

If we already knwow that $X$ is $T_3$, $X$ is compact iff every embedding $e$ of $X$ into a Hausdorf space $Y$ has a closed image $e[X] \subset Y$. (in general such a space is called H-closed, and H-closed $T_3$ spaces are compact).