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I've been struggling with providing examples of the following:

1) A continuous function $f$ and a connected set $E$ such that $f^{-1}(E)$ is not connected

2) A continuous function $g$ and a compact set $K$ such that $f^{-1}(K)$ is not compact

Alex C
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3 Answers3

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Let $f(x) = \sin x$, and let $E = [0,1]$. Then $E$ is compact and connected, but $f^{-1}(E)$ is the disjoint union of infinitely many closed intervals, and is therefore neither compact nor connected.

Jim Belk
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  • He mentions inverse function, so sin is no example. – peter Jan 31 '24 at 09:52
  • @peter The notation $f^{-1}(E)$ refers to the inverse image of a set $E$ under the function $f$. The function $f$ itself does not need to be invertible. – Jim Belk Jan 31 '24 at 09:56
  • what you say is true but doesnt touch the fact that the headline mentions an inverse function. – peter Jan 31 '24 at 10:00
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Take any space $X$ which is not connected and not compact. For example, you could think of $\mathbf R - \{0\}$. Map this to a topological space consisting of one point. [What properties does such a space have?]

Dylan Moreland
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For an example of an invertible function for the second part, map $[0,2\pi)$ to the unit circle in $\Bbb R^2$ via $t\rightarrow(\cos t, \sin t)$. $f$ is continuous, the unit circle $S$ in $\Bbb R^2$ is compact, but $f^{-1}(S)$ is not.

For the other example, take the space to be $[0,1)\cup[2,3]$ and the map to be $$ f(x)=\cases{x,& $x\in[0,1)$\cr x-1,&$x\in [2,3]$ } $$ Then $f$ is continuous, invertible, $[0,2]$ is connected, but $f^{-1}([0,2])$ is not. (This actually furnishes an example for both parts.)

David Mitra
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