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Explain precisely why a continuous surjection $f: [0,1] \mapsto [1, \infty)$ does / does not exist

I am convinced that such function does not exist. My agument is that - according to the Weierstrass Theorem on continuous functions, if a function is continuous on $[a,b]$ then there are two points in the closed interval that map into the supremum and the infimum of the function, respectively. Since the function is continuous on a closed interval and it does not take a maximal value - it does not exist.

Is my reasoning correct?

Aemilius
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1 Answers1

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Your argument works. Here's an easier one: $[0, 1]$ is compact, and the image of a compact set under a continuous map is compact. $[0, \infty)$ is not compact.

Duncan Ramage
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  • I don't think this is easier, unless you assume the results on compactness as black boxes. – Andrés E. Caicedo Jan 10 '18 at 17:55
  • @AndrésE.Caicedo Why? – Duncan Ramage Jan 10 '18 at 17:58
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    (+1) See also this very similar question (and answer): https://math.stackexchange.com/q/2598681 – Fabio Lucchini Jan 10 '18 at 18:13
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    Of course it is easier. This just uses that the continuous image of a compact space is compact. –  Jan 10 '18 at 18:57
  • @Math_QED No, it also uses the concept of compactness, the fact that $[0,1] $ is compact and $[0,\infty) $ is not. How do you prove that continuous images of compact sets are compact? Probably you use the topological characterization of continuity (rather than the $\epsilon $-$\delta $ one), so you also have to establish that. Do you argue for arbitrary topological spaces (so you need the notion of topological space) or explicitly for the case at hand, thus having to in addition translate the nice clean argument you have in mind into a messier one? Etc. But the key thing, really, is ... – Andrés E. Caicedo Jan 11 '18 at 02:38
  • @Math_QED ... that none of these external concepts are really needed here, so this suggested argument is creating a sophisticated abstract apparatus for no real reason (or perhaps creating a poor version of it to stick to the concrete setting on this question). And absorbing the required wealth of new notions (even in watered-down form and only superficially) is a harder task than verifying some routine results familiar from calculus. – Andrés E. Caicedo Jan 11 '18 at 02:42
  • That continuous functions push forward compact sets is a routine result familiar from calculus. At least, the level of calculus were one can quote the Weierstrass theorem. – Duncan Ramage Jan 11 '18 at 02:56
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    All the results you mention are standard in any real analysis course that handles metric spaces. –  Jan 11 '18 at 08:48
  • The question could be posed to students who have only seen the notion of convergence for sequences of real numbers, together with the least upper bound axiom, with continuity introduced via "$f(\lim_{n \to \infty} x_n) = \lim_{n\to \infty} f(x_n)$". As such, an answer that requires someone to be conversant with the notion of compactness, or indeed the definition of a metric space, is needlessly elaborate. – krm2233 Aug 04 '23 at 19:57