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If $A$ is compact, is then $f(A)$ compact?

The answer here by David Mitra uses $f^{-1}$, however we only know $f$ is continuous in its domain, so how do we come up with the inverse?

Its not mentioned to be strictly montone either. I do think it's a stupid question but I really want to know.

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    $\begingroup$ If $A$ is a set and $f:X\to Y$ a function, then $f^{-1}(A)$ is always defined and is by definition $\{x\in X\mid f(x)\in A\}$. $\endgroup$
    – Surb
    Aug 1 '20 at 7:42
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For any function $f: X \to Y$ and any set $A \subseteq Y$ we define $f^{-1}(A)$ as $\{x \in X: f(x) \in A\}$. This is called the inverse image of $A$ under $f$.

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  • $\begingroup$ Thanks, I got confused $\endgroup$
    – user732848
    Aug 1 '20 at 7:44
  • $\begingroup$ @Shamim It may in fact be confusing. The symbol $f^{-1}$ has two interpretations: (1) For each subset $A \subset Y$, $f^{-1}(A)$ denotes the inverse image (also called preimage) of $A$ which is a subset of $X$. Thus you may regard $f^{-1}$ as function from the power set of $Y$ to the power set of $X$. (2) If $f : X \to Y$ is bijective, then it is standard to write $f^{-1} : Y \to X$ for the inverse bijection. Usually it is absolutely clear from the context which interpretation is meant. $\endgroup$
    – Paul Frost
    Aug 1 '20 at 8:08
  • $\begingroup$ Yes, I knew that definition beforehand but forgot right then, so I got confused. $\endgroup$
    – user732848
    Aug 1 '20 at 9:04
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In your given link, $f^{-1}$ is only used just for preimage of the set $A$. So, in this case, $f^{-1}$ need to be just a relation! You can say , bijective is necessary, if $f^{-1}$ is used as $f^{-1} (a) $ for $a\in Im(f)$, which is a inverse function of $f$.

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  • $\begingroup$ Thanks for the answer $\endgroup$
    – user732848
    Aug 1 '20 at 7:58

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