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.

  • 2
    $\begingroup$ By the wording of the question, it seems that you're not supposed to just find one function on a non-compact domain with the specified property; you have to show that such a function exists for any non-compact domain. $\endgroup$
    – florence
    Jun 18 '17 at 19:28

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.


Let $E$ be a non-compact subset of $\mathbb{R}$. Then we either have that $E$ is not bounded or $E$ is not closed. If the former holds, note that

  1. $f(x) = x^2$ is unbounded on $E$.
  2. $f(x) = \arctan(x)$ is bounded and has no maximum.
  3. The function from 1. also is not uniformly continuous on $E$.

If $E$ is not closed find $ p \notin E$ in its closure. Play with functions like $f(x) = \frac{1}{|x-p|}$ to construct functions as required.


As you've noted, a subset of $\mathbb{R}^n$ is compact $\iff$ it is closed and bounded. So if $E$ is not compact, then either it is unbounded, or it is bounded but not closed.

If $E$ is unbounded, something like your $f(x) = \sqrt{x}$ example will work. However, you need to define it piecewise to ensure that it isn't taking invalid input from $E$ (negative numbers). You could also choose any continuous function $g$ on $\mathbb{R}$ such that $\displaystyle \lim_{x \rightarrow - \infty} g(x) = -\infty \text{ and } \lim_{x \rightarrow \infty} g(x) = \infty$ or vice-versa.

Now let's consider when $E$ is bounded but not closed. In this case, we can take advantage of functions like $\displaystyle \frac{1}{x}$. If $E$ were, say, $(0, 1]$, then $f(x) = \displaystyle \frac{1}{x}$ itself would work since $\displaystyle \lim_{x \rightarrow 0^+} f(x) = \infty$. Think about how to generalize this idea to arbitrary bounded-but-not-closed sets.

It is also worth thinking about whether a function $f:E \rightarrow \mathbb{R}$ is necessarily bounded when $E$ is compact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.