Let $I$ be a closed real interval.

Let $f \colon I \longrightarrow \mathbb{R}$ be a real continuous function so it has a global maximum point.

If I say “let $x_0$ be a point of global maximum…”, my question is: am I using axiom of choice?


marked as duplicate by Asaf Karagila axiom-of-choice Dec 6 '17 at 15:44

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No, you're not. The axiom of choice isn't relevant for making one single choice.

The axiom of choice only becomes relevant when you have to make infinitely many choices. Some times, when you have to make infinitely many choices, you can give a general rule for how to choose, and apply it to all those choices simultaneosly. The axiom of choice states that even when you cannot find such a rule, it is still possible to make all those infinitely many choices in one go.

  • $\begingroup$ Does it require the axiom of choice to prove that real-valued functions have global maxima on closed intervals, though? $\endgroup$ – Mees de Vries Dec 6 '17 at 15:29
  • $\begingroup$ @MeesdeVries It might, now that you say it. Well, at least countable choice. $\endgroup$ – Arthur Dec 6 '17 at 15:30
  • $\begingroup$ not to prove but just to define it $\endgroup$ – Matey Math Dec 6 '17 at 15:30

No. The set of all points where $f$ attains a global maximum is a closed subset of $I$ and therefore it is a compact set. So, it has a minimum. Therefore, you could, for instance, choose the minimum among the ponts at which $f$ attains a global maximum.

So, even if you had infinitely many functions (possibly with distinct domains) the axiom of choice would not be required in order to choose a point of global maximum for each of them.

  • 1
    $\begingroup$ This becomes more interesting if we say that the funtions don't have to be continuous, just that they have a maximum on $I$. Then your last paragraph doesn't apply any more, and we would need the axiom of chioce. $\endgroup$ – Arthur Dec 6 '17 at 15:34
  • $\begingroup$ @Arthur Yes, of course. $\endgroup$ – José Carlos Santos Dec 6 '17 at 15:36

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