Timeline for $X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2023 at 19:31 | comment | added | Brian M. Scott | @CrashBandicoot: You’re very welcome! | |
| Jan 16, 2023 at 7:43 | comment | added | Crash Bandicoot | Even if belated, I appreciated a lot your answer! Thank you very much! | |
| Jan 15, 2023 at 3:50 | comment | added | Brian M. Scott | @CrashBandicoot: This is a very belated answer, since I only just realized that you never did ask the question separately. The question was in fact asked and answered somewhat earlier: we cannot always construct such a function. Indeed, there are even non-compact spaces on which every continuous real-valued function is constant. | |
| Apr 8, 2021 at 21:10 | comment | added | Crash Bandicoot | @BrianMScott sorry I did not specify the question I was looking for. I was looking for an example where $X$ is a generic non-compact topological space. We know that if $X$ is a noncompact metric space then there exists a function whose image is not closed. I was wondering if we can always construct such a function also in the case $X$ is a generic topological space. But I think this deserves a different question. Thanks anyways! | |
| Apr 8, 2021 at 18:09 | comment | added | Brian M. Scott | @LorenzoLotto: The identity function from $(0,1)$ to $\Bbb R$ is an example. For that matter, so is the identity function from $\Bbb R$ to $\Bbb R$. Any injection from $\Bbb N$ to $\Bbb R$ whose range does not have a smallest or largest element is another example. | |
| Apr 8, 2021 at 15:45 | comment | added | Crash Bandicoot | @BrianM.Scott do you have an example of a function from a non-compact topological space to the real line that does not attain its max/min? | |
| Jul 21, 2020 at 18:42 | comment | added | hteica | @BrianM.Scott I see, I forgot this crucial fact. | |
| Jul 20, 2020 at 17:13 | comment | added | Brian M. Scott | @hteica: The default assumption is that spaces are not empty. | |
| Jul 20, 2020 at 13:06 | comment | added | hteica | This has been bothering me for a while, to see that $f[X]$ is bounded implies $f[X]$ has both a supremum and infimum, we need the fact that $f[X]$ is nonempty, right? but if we take $X$ to be the empty set, then it is compact and $f$ is then the empty set, so it is continuous, so that $f[X]$ is also an empty set. Thanks. | |
| Oct 19, 2015 at 3:39 | comment | added | Brian M. Scott | @grayQuant: The proof actually used only the compactness of $X$; the hypothesis that $X$ is metric isn’t needed. If $X$ is not compact, however, there may be continuous real-valued functions on $X$ that do not attain at least one extremum. | |
| Oct 19, 2015 at 3:35 | comment | added | grayQuant | Nice proof, how does this generalize to X is an arbitrary topological space? will the same proof hold? | |
| Feb 15, 2012 at 20:49 | vote | accept | Emir | ||
| Feb 15, 2012 at 7:29 | history | answered | Brian M. Scott | CC BY-SA 3.0 |