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I need help to show that a set E of real numbers is closed and bounded if and only if every continuous function on E takes a maximum value.

I can use The Intermediate Value Theorem to prove if E is closed and bounded then every continuous function on E takes a maximum value. But converse part I do not know how to prove. Please help. Thanks

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    If $E$ isn't closed, let $a$ be a limit point of $E$ not in $E$, and consider the function $x \mapsto 1/(x-a)$ and $x \mapsto -1/(x-a)$. If $E$ is unbounded, then consider $x \mapsto x$ and $x \mapsto - x$. – 2017-02-05
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    I guess the theorem you wanted to use was the extreme value theorem, not the intermediate value theorem. – 2017-02-05
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    One direction is known as the extreme value theorem. A proof can be found [here](http://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min). The other direction has a proof that can be found [here](http://math.stackexchange.com/questions/114123/if-every-continuous-function-attains-its-maximum-then-the-metric-space-is-comp) – 2017-02-05

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