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I am looking for a function $f: R\rightarrow R$ and a sequentially compact $K\subset R$ such that the inverse $f^{-1}(K)$ is not sequentially compact.

I decided to choose $f(x) = sin(x)$, but I'm not sure what $K$ could be such that $f^{-1}(K)$ is not sequentially compact.

Are there any ideas about what $K$ can be? I'm having a hard time grasping the concept of sequentially compact.

I'm thinking $K=[-\pi/2,\pi/2]$

samsonite
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  • That works since $f^{-1}[-\pi/2, \pi/2] = \mathbb{R}$, which is not sequentially compact. You could take any interval $[a,b]$ with $[-1,1] \subset [a,b]$ and your function. – Ethan Alwaise Nov 24 '15 at 05:06
  • In this case, any set which has nonempty preimage has unbounded preimage. – tomasz Nov 24 '15 at 05:32

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You can get even simpler than this, when in doubt look at the extreme cases. Define $f(x)=0$ to be the constant function. Then {0} is sequentially compact, but $f^{-1} (\{ 0\})=\mathbb R$, which isnt.

Alan
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