# Continuous function with no max in closed and bounded set

Okay guys, I have this question that troubles me a lot. Is there an example of a function that is continuous on a closed and bounded set but achieves no maxima?

My take is, that apparently cannot be in Euclidean space. I think of bounded sequences, L-inf'ty, where seqs are bounded and closed (all limits contained) but my puzzle is then why not to have a maximum (i.e. subseqs will not also converge to the limit contained)? Plus that I cannot come up with an example of such a function.

Thanks a lot.

P.S. THis is my first post and I am a rookie in math, so I apologise if I don't express something very clearly.

Edit: Thanks a lot for your super quick responses! I forgot to mention that I consider a sup metric in L-inf'ty space I mention. Of course this is just an example.

• @DougM Not all closed and bounded sets are compact though. Sep 21 at 1:13
• Since you put the metric-spaces tag, consider any infinite, discrete metric space. Then any map to $\Bbb{R}$, including unbounded ones, will be continuous, despite the the metric space being bounded (and closed within itself, as always). Sep 21 at 1:15
• It's easy to cook up an example inside $\Bbb{Q}$. Perhaps you should tell us a bit more about the context: general metric spaces? Hilbert spaces? Or what? Sep 21 at 1:15
• @GiorgosGiapitzakis that is correct, and if I am not mistaken, a continuous mapping of a set that is closed and bounded and not compact, does not necessarily have a maximum. Sep 21 at 1:19
• Awesome community guys! Had not tried it before. Very motivating to keep up. Sep 21 at 1:32

You can easily verify that $$\mathbb{R}$$ equiped with the metric $$d: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ defined as $$d(x,y) = \begin{cases}0,\; x=y \\1,\; \text{otherwise} \end{cases}$$ is indeed a metric space. In particular, every set is both open and closed. Now consider the function $$f: ([-1,1], d) \to (\mathbb{R}, d_{\text{euc}})$$ given by $$f(x) = \begin{cases} \frac{1}{x}, \; x \neq 0 \\ 0, \; \text{otherwise} \end{cases}$$ where $$d_\text{euc}$$ is the usual metric on $$\mathbb{R}$$. Now $$f$$ is continuous since the inverse image of every open set is open and $$[-1,1]$$ is closed and bounded but $$f$$ is unbounded (and in particular achieves no maximum value).
• @peter if the answer helped, consider upvoting it. $+1$ Sep 21 at 1:39
I'll add an example along your original thoughts: Consider, for every $$n\in \mathbb{N}$$, the sequence $$X_n\in \ell_\infty$$ given by $$X_n=(0, \ldots, 0, 1-\frac{1}{n}, 0, \ldots),$$ where the only nonzero term is in the $$n$$-th place. Now let your set be $$E=\{ X_1, X_2,\ldots\}\subset \ell^\infty.$$ Clearly $$E$$ is bounded since $$\| X_n\|_\infty =1-1/n \leq 1$$. It's also closed, since the only possible accumulation point would have all entries 0 but at the same time $$\ell_\infty$$ norm 1 which is clearly not possible.
Then $$f:E\to \mathbb{R}$$ given by $$X\mapsto \| X\|_\infty$$ doesn't attain a maximum.