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I know this is part of the Extreme Value Theorem but I want to tackle this one bit first, focusing on the maximum case.

For a function $f$ continuous over interval $[a,b]$, it has a max value over this interval.

I define the max like so: $\exists d \in [a, b] : \forall x \in [a, b], f(x) \leq f(d)$

One way to begin tackling this is to prove by contradiction by showing the negation is false.

$\forall d \in [a, b] , \exists x \in [a, b]: f(x) > f(d)$

In other words if there is no maximum then it means for any $d$ we choose, there's a value of $x$ where $f(x)$ is even greater.

The usual proofs I see just throw a bunch of set theory and notation at the problem that I don't understand, involving things like subsequences or "compactness" for some reason. I don't understand the use, purpose, or motivation for these approaches.

Is the general idea to show that the negation implies the function going up to infinity, which somehow contradicts our assumption of continuity?

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  • $\begingroup$ It is absolutely necessary that you understand what is going on concerning this problem. As long as you as you are thinking along the lines of the paragraph "The usual proofs $\ldots" in your question you are not yet there. Read the following answers carefullly! $\endgroup$ – Christian Blatter Mar 21 '18 at 16:01
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The negation says: "it is not bounded". So yeah the idea is to use that and reach a contradiction.

Without compactness you can't do it. Compact=bounded and closed. If it is open say on the right $[a,b)$ then the function can explode in b and still be continuous. Look at $1/x$ on $[-1,0)$.

If it is not bounded then just take literally any continuous function that goes to infinity, say a non constant polynomial...

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  • $\begingroup$ What is "bounded and closed"? I mean I know something like $[a$ is closed and $(a$ is open but I don't know is relevance here, or what's considered "bounded" and whether or not that's actually proving anything or just kicking the can to some other concept $\endgroup$ – user525966 Mar 21 '18 at 15:38
  • $\begingroup$ I don't think anything in a proof can be there as an option. Everything is there for a reason. bounded means that [a,b] for any pair of numbers in [a,b] their difference is finite (or rather, you can't find two sequences $a_n$ and $b_n$ in [a,b] such that lim$a_n-b_n=\infty$). closed means that you can't find a sequence $a_n\in [a,b]$ such that it's limit is outside of [a,b] $\endgroup$ – John Cataldo Mar 21 '18 at 15:42
  • $\begingroup$ $\forall d \in [a, b] , \exists x \in [a, b]: f(x) > f(d)$ is the same as saying it is unbounded and closed? $\endgroup$ – user525966 Mar 21 '18 at 19:10
  • $\begingroup$ just unbounded. "Closed" is about the set [a,b] here. An equivalent definition of closed is that $[a,b]^c=(-\infty,a)\cup(b,\infty)$ is open. Just google it. $\endgroup$ – John Cataldo Mar 21 '18 at 19:14
  • $\begingroup$ Yes I refer to "[a,b]" with closed but then why do you mean "just unbounded"? $\endgroup$ – user525966 Mar 21 '18 at 19:16
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There are two main hypotheses for this result: that the function is continuous, as you have noted, and that the domain which you consider is an interval, that is, a compact (bounded and closed) set.

Consider the continuous function $f(x)=x$ over $[0,1)$, that is, the set of $x$ such that $0 \leq x < 1$. You cannot find a point $d$ for which $f$ is maximal over this set : there will always be a point (closer to 1) where $f$ is larger.

The same reasoning works for the closed but unbounded set $\mathbb{R}$.

This is the reason why proving the above statement requires you to use some result relating to compactness and the fact that sequences will converge within a closed interval.

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  • $\begingroup$ Why can't I say what the max value is for $[0, 1)$? What's the difference between bounded and closed/open? Are all $4$ permutations possible? Bounded and open? Bounded and closed? Unbounded and open? Unbounded and closed? $\endgroup$ – user525966 Mar 21 '18 at 15:40
  • $\begingroup$ I do not understand your comment. Over $[0,1)$, the maximum value of $f(x)=x$ is never reached. For any $d$ as close as you will from $1$, the value $f(d+\frac{1-d}{2})$ will be larger than $f(d)$. $\endgroup$ – Joce Mar 21 '18 at 15:41
  • $\begingroup$ Well, you are now asking for basic definitions, which you will find in your textbook. Your original question was relevant, "why do we need this?", I think you can understand the problem with $[0,1)$, and be convinced that you need now to follow the usual way of learning what mathematicians have done to address the issue. $\endgroup$ – Joce Mar 21 '18 at 15:44
  • $\begingroup$ (I don't have a textbook, this is self-learning.) Is the lack of a max "by definition" of a set? I still can't tell if this is a "bound" issue or a "open/closed" issue $\endgroup$ – user525966 Mar 21 '18 at 15:46
  • $\begingroup$ You need both, as I state in my answer. Wikipedia can make up for a textbook, but I'd recommend getting one. $\endgroup$ – Joce Mar 21 '18 at 15:48

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