Proof of Extreme Value Theorem. Modus operandi. (S. Abbott pp 115 t4.4.3)

Question

By pp 115 Abbott Theorem 4.4.2, we know that $f([a, b])$ is bounded above and below.
$f([a, b])$ contains [a,b] ergo it is clearly nonempty).
By the agency of the Axiom of Completeness, it has a supremum. Name it $s$.
Now for each natural number n, the number $s - 1/n$ is not an upper bound.
Thus there is a number in $f([a, b])$, call it $f(x_n)$, such that $s - 1/n < f(x_n) ≤ s.$
Ergo the sequence $\color{seagreen}{f(x_n) \text{ converges to } s.}$

1. Is the last sentence by dint of Squeeze Theorem? If not, why does $f(x_n)$ converges to s?

$\{ x_n \} $ is a sequence in $[a, b]$, so by the Bolzano–Weierstrass Theorem,
$\{ x_n \} $ has a subsequence $\{x_{n_k} \}$ that converges to an element in [a, b], call it $x_0$.

Then since f is continuous, the image of this subsequence, $f( x_{n(k)} )$ must converge to $f(x_0)$.
But since $f( x_{n(k)} )$ is a subsequence of $\color{seagreen}{\text{a sequence that converges to } s}$, $f( x_n ) $ must converge to s. Thus $s = f(x_0)$. So the supremum is a function value.

Now note that since f is continuous, so is –f.

Since –f has a maximum (by above) f has a minimum at the same point.

2. Is the nub to the proof $\max f = \sup f $? How can we presage this before proving?

3. I acquiesce to the proof's steps, but I don't understand the modus operandi?
How can we presage the need of subsequences? Why chagrin about subsequences?

Answer 1

1. It is by the squeeze theorem on the "lower" sequence $a_n=s-1/n$ and the constant "upper" sequence $b_n=s.$ You then have $a_n < f(x_n) \le b_n$ so the squeeze theorem applies, since the length $b_n-a_n=1/n\to 0.$ This then gives $f(x_n) \to s.$

3. This is a construction of an initial sequence $x_n$ where each term is somewhere in $[a,b]$. But one wants a single element $x_0$ of $[a,b]$ for which $f(x_0)=s.$ That is the reason for using a subsequence of the already constructed sequence $x_n$. Any subsequence of $x_n$ will have its $f$ values converging to $s$.

Then since the points $x_n$ are in the bounded interval $[a,b]$ one can find a limit point $x_0$ of these $x_n$, which is a point for which any neighborhood contains $x_n$ for infinitely many subscripts $n$. Now, there is no way around constructing a subsequence of the $x_n$ which converge to $x_0$, and then one shows that $f(x_0)=s$ as in the quoted argument.

In summary: The first sequence $x_n$ is to make the $f$ values approach $s$, while the need for a subsequence of that is so one can have also the $x_n$ values approach some $x_0$ in $[a,b]$. Note there may be a way to do this proof without sequences at all, but this sequence approach isn't very complicated. I'd also be curious to see a non-sequence proof.

To address question 2: The theorem is saying that in this situation: $\max f(x)=\sup f(x).$

One might "presage" this before proving it, after looking at several cases. However one would not know it for sure until a proof was obtained or else looked at from some source. Often one is given an exercise to prove something, and doesn't really know in advance that it must be true in the first place. It's of course always good to try for a counterexample first in terms of understanding what's going on.

Some more explanation: (for the OP, who requested that some things be explained in the answer part, rather than in comments.)

By "This is a construction..." I was referring to the quoted part of the proof in the posted question, or basically to the sequence and subsequence constructed in the proof.

By saying "there is no way around" I only meant to imply that, once one had gotten that far, it would seem clear that one should next construct the subsequence. "no way around" is not meant to mean that there is no other way, only that the subsequence step is an obvious way that works, so no need for other ways to do it.