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true or false: if the domain of a function is compact then the range is compact. my guess is it's True, but I keep thinking this can only be true if the function is continuous. if this is not true can you please explain or give a counterexample?

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Not true in general, you can map $[0,1]$ to $\Bbb R$ because both have the same cardinality.

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It's true that the image of a compact space is compact under a continuous map.

Let $X$ be compact and $Y = f(x)$ be spaces where $f: X \rightarrow Y$ is continuous. Let $W$ be an open cover of $f(X)$ by open sets in $Y$. Since $f$ is continuous $f^{-1}(W)$ is open in $X \ \forall w \in W$. Then the set $\{f^{-1}(W): w \in W \}$ is an open cover of $X$. Then since $X$ is compact it has a finite subcover $\{f^{-1}(w_1), f^{-1}(w_2), \dots f^{-1}(w_r) \}$ therefore $\{w_1, w_2, \dots , w_r\}$ is a finite subcover of $f(x) = Y$ so $Y$ is compact.

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