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Is

$$ \sup_{x\in S} f(x) \equiv \max_{x\in S} f(x) $$

And is the correct definition of a maximum of a function the following: $$ \max_{x\in S} f(x):= y\quad \text{such that}\ \exists x\in S,y=f(x)\wedge\forall x\in S, f(x)\le y $$

Ben
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    Definition of max is correct, but supremum and maximum may not be the same (because on certain sets, max doesn't exist). Consider $f(x) = x$ on $S = (0,1)$. What's the max of $f$? It can't be $1$ since $1 \notin S$. We see $f$ has no maximum. Instead, the supremum of $f$ is $1$ (since supremum doesn't have to be in the set) – Sam Nov 12 '22 at 16:48
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    The max is always reached by the function, not necessarily the sup. – Lelouch Nov 12 '22 at 16:49
  • @Sam ah that makes sense. Would the supremum, maximum equality hold if $S$ was closed...? – Ben Nov 12 '22 at 17:19
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    @Ben You would also need continuity (otherwise, you could take Sam's example and also say that $f(0)=f(1)=0$). And you need the interval to be bounded (otherwise look at $-1/x$ on $[1,\infty)$). But there is a theorem (usually seen in the first few weeks of Calculus 1) that a continuous function on a closed and bounded interval does have a maximum. – Teepeemm Nov 12 '22 at 18:23
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    @Ben, being continuous on a closed and bounded interval would guarantee a maximum like Teepeemm said, but that too is a strong condition. A simpler condition would be $f(S)$ i.e the image of $S$ being closed and bounded. For example $f:(0,1) \cup (2,3) \to\mathbb{R}$ such that $f(x) = 2$ for $0 < x < 1$ and $f(x)= 3$ for $ 2 < x <2.5 $ and $f(x) = 7$ for $2.5 \leq x < 3$ has a maximum, even though $S$ is neither closed and bounded, nor is an interval, nor is $f$ continuous. All you care about is if the image of $f$ is closed and bounded. (Also you need $f(S)$ non empty) – Sam Nov 12 '22 at 19:25

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