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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.

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    $\begingroup$ @DougM Not all closed and bounded sets are compact though. $\endgroup$ Sep 21 at 1:13
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    $\begingroup$ 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). $\endgroup$ Sep 21 at 1:15
  • $\begingroup$ 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? $\endgroup$
    – Rob Arthan
    Sep 21 at 1:15
  • $\begingroup$ @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. $\endgroup$
    – Doug M
    Sep 21 at 1:19
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    $\begingroup$ Awesome community guys! Had not tried it before. Very motivating to keep up. $\endgroup$
    – Peter21
    Sep 21 at 1:32
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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).

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  • $\begingroup$ That is great! Thank you a lot! $\endgroup$
    – Peter21
    Sep 21 at 1:30
  • $\begingroup$ @peter if the answer helped, consider upvoting it. $+1$ $\endgroup$
    – Clayton
    Sep 21 at 1:39
  • $\begingroup$ Giorgos I changed the accepted answer just because Jose's maps more closely the content of the question, for others that might have same question. Also happy, to "upvote" as Clayton suggests provided that I figure out how. :P $\endgroup$
    – Peter21
    Sep 21 at 2:43
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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.

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  • $\begingroup$ Many thanks Jose, now I have an illustration of exactly what was puzzling me. $\endgroup$
    – Peter21
    Sep 21 at 2:39

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