# Proving an inverse image is sequentially compact

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.
• what's $\overline{B}(0, 1)$? also I am not familiar with metric spaces – user641672 Feb 12 '19 at 4:13
• 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. – Matematleta Feb 12 '19 at 4:14