Let $f$ be a continuous function. I know that if $A$ is compact then $f(A)$ is compact but is $f^{-1}(A)$ also compact? I believe it is not but how can I prove it by a counter example?
2026-04-26 08:39:24.1777192764
If $A$ is compact then $f^{-1}(A)$ compact?
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Consider $f:\mathbb R\to\mathbb R $ defined by $f (x)=1$. The set $\{1\} $ is compact, but $f^{-1}(\{1\})=\mathbb R $ is not.