Let $E\subset\mathbb{R}$ a non-compact set. Show that
i) Exists a continuous function in $E$ that is not bounded.
ii) Exist a continous and bounded function in $E$ that doesn't have
maximum.iii) If $E$ is bounded then, exists a continuous function in $E$ that
is not uniformly continuous
I started writting one definition that I have doubt
Def: A set $E\subset \mathbb{R}$ is compact if and only if it is
closed and bounded.
If a set is non-compact then I can say that that it is not bounded or not closed?
i) $f(x)=\sqrt{x}$ is a continous function and it is not bounded.
ii) From the doubt that I have in definition, if a set is non-compact and it is bounded then it is not closed so doesn't have a maximum.
iii) I don't know how to find a example.
Best Answer
Either $E$ is not closed or not bounded (or both).
(i) If $E$ is not closed, let $a \in \mathbb R$ be a limit point of $E$ that is not contained in $E$. Then $\displaystyle f:E \to \mathbb R, \quad f(x)= \frac{1}{x-a}$ is an unbounded function.
If, on the other hand, $E$ is not bounded, then $f:E \to \mathbb R, \quad f(x)=x$ is unbounded.
(ii) If $E$ is not closed, choose $a \in \mathbb R$ as above. Then (for example) $f(x)=e^{-(x-a)^2}$ works. (because $e^{-x^2}$ is a "bell" curve, continuous, bounded and only have one maximum)
If $E$ is not bounded from above, $f(x)= \arctan x$ works. If $E$ is not bounded from below, then you can take $f(x)=-\arctan x$.
(iii) $\displaystyle \ f(x)= \frac{1}{x-a}$ works for this case, too.