I am studying Real Analysis and I had a exam last week. In that test there was the question below:

Let $U \subset \mathbb{R}^n$ a non compact set. Show that there exists continuous function $f: U \to \mathbb{R}$, such that $f$ is unbounded.

So I tried to show that if every $f:U \to \mathbb{R}$ continuous is bounded then $U$ is compact.

I have proved that $U$ is bounded, but how can I prove that $U$ is closed??? It's just for curiosity because the test was last week.

  • $\begingroup$ What if $U=\mathbb R$? Can you think of an unbounded continuous function $f:\mathbb R\to\mathbb R$? What's your favourite continuous function on $\mathbb R$? $\endgroup$
    – Dave
    Sep 21 '19 at 2:48
  • $\begingroup$ This is trivial, it is the identity. I have thought this but... $\endgroup$
    – Joãonani
    Sep 21 '19 at 2:50

HINT: If $a\in\Bbb R^n$ is in the closure of $U$ but not in $U$, consider the function $f(x) = 1/\|x-a\|$.


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