Suppose $f : \mathbb{R}^{n} \rightarrow \mathbb{R}$ is continuous and $f(u) \geq ||u||$ for every $u \in \mathbb{R}^{n}$. Prove $f^{-1}([0, 1])$ is sequentially compact.

A set $S \subseteq \mathbb{R}^{n} $ is sequentially compact if every sequence in $A$ has a subsequence that converges to a point in $A$.

My thoughts:

I am not too sure about how to approach this problem. I think the best way to go is to just show that the set is closed and bounded. I think there's a theorem about inverse images being sequentially compact if the original set is compact, but I did not learn that one. I did learn that images of sequentially compact sets are compact, though. This is sort of the other direction.

I also don't know where the inequality comes into play. I feel like it's sort of arbitrary and I don't know where I should use it.

Any help is appreciated.


If $u\in f^{-1}([0,1])$ then $f(u)\in [0,1]$ so $\|u\|\le 1$ which implies that $u\in \overline B(0,1)$ which is compact. And as $f^{-1}([0,1])$ is closed because $f$ is continuous, we conclude that it is compact, too, hence sequentially compact because $\mathbb R^n$ is a metric space.

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  • $\begingroup$ what's $\overline{B}(0, 1)$? also I am not familiar with metric spaces $\endgroup$ – user641672 Feb 12 '19 at 4:13
  • $\begingroup$ The closed unit ball.All you really need to know is that $f^{-1}([0,1]$ is closed and bounded and so is sequentially compact by Bolzano-Weierstrass. And this is what I proved above. $\endgroup$ – Matematleta Feb 12 '19 at 4:14

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