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I was asked to prove the following statement:

Let $K \subseteq R^n$. show that $K$ is compact (meaning closed and bounded) if and only if every continuous function is bounded on $K$.

What I did:

Suppose $K$ is not bounded, and so, it is not compact. Then the function $\sum |x_i|$ is a continuous unbounded function on $K$. Via contrapositive, this shows that if every function is bounded, then $K$ is also bounded.

What I need help with:

Assume $K$ is not closed. I need to find a continuous and unbounded function on $K$.

that will prove that if every continuous function is bounded on $K$, then $K$ is compact.

after that, i still need to show that if $K$ is compact then every continuous function $f: K \to \mathbb R$ is bounded.

Would someone point me in the right direction?

Clarification: it's not homework. I am preparing for an exam.

Rushabh Mehta
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Oria Gruber
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    Do you know how you can use the distance to characterise the closure of a (nonempty) set? – Daniel Fischer Jun 21 '14 at 21:01
  • $\alpha$ is a point in the closure of $K$ if for every $\epsilon >0$ there is an $x \in K$ such that $|\alpha - x|< \epsilon$. is that what you mean Daniel? – Oria Gruber Jun 21 '14 at 21:04
  • http://math.stackexchange.com/questions/449358/true-or-not-compact-iff-every-continuous-function-is-bounded – Riccardo Jun 21 '14 at 21:05
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    haa!! suppose $K$ is not closed, and $\alpha$ is a limit of $K$ while not being in $K$. Then $f:K \to \mathbb R$, $f(x)=\frac{1}{|x-\alpha |}$ is not bounded! so indeed, if every continuous real valued function is bounded, then $K$ is compact. – Oria Gruber Jun 21 '14 at 21:08
  • Bingo, @OriaGruber! – Daniel Fischer Jun 21 '14 at 21:09
  • How can we show the other direction? That compactness of $K$ implies every continuous function is bounded? Suppose $f: K \to \mathbb R$ is continuous and not bounded. What does it say on $K$? – Oria Gruber Jun 21 '14 at 21:14
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    The continuous image of a compact set is compact. – David Mitra Jun 21 '14 at 21:19

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If $K$ is not closed let $a\in \overline{K}\setminus K.$ Let $d_E$ denote the Euclidean metric then the function $H:K\to\mathbb{R} $$$H(t) =\frac{1}{d_E (t,a)}$$ is continouos and not bounded.