I am self-teaching from Pugh's Real Analysis text. I am trying to grasp the proof of Theorem 23 from Chapter 1 which states
A continuous function $f$ defind on an interval $[a,b]$ takes on absolute minimum and absolute maximum values: For some $x_0, x_1 \in [a,b]$ and for all $x \in [a,b]$ we have
$$f(x_0) \leq f(x) \leq f(x_1)$$
His proof:
Let $M = \sup f(t)$ as $t$ varies in $[a,b]$. BY Theorem 22, $M$ exists. Consider the set $$X = \{x \in [a,b] : \sup V_x < M\}$$
where $V_x$ is the set of values of $f(t)$ as $t$ varies on $[a,x]$, i.e.
$$V_x = \{ y \in \mathbb{R} : \text{ for some } t \in [a,x] \text{ we have } y = f(t)\}$$
Case 1. $f(a) = M$. Then $f$ takes on a maximum at $a$ and the theorem is proved.Case 2. $f(a) < M$. Then $X \neq \emptyset$ and we can consider the least upper bound of $X$, say $c$. If $f(c) < M$, we can choose $\epsilon > 0$ with $\epsilon < M – f(c)$. By continuity at $c$, there exists a $\delta > 0$ such that $|t-c| < \delta$ implies $|f(t) – f(c)| < \epsilon$. Thus $\sup V_c < M$. If $c < b$ this implies that there exist points $t$ to the right of $c$ at which $\sup V_t < M$, contrary to the fact that $c$ is an upper bound of such points. Therefore, $c = b$, which implies that $M < M$, a contradiction. Having arrived at a contradiction from the supposition that $f(c) < M$, we duly conclude that $f(c) = M$, so $f$ assumes a maximum at $c$. The situation with minima is similar.
I am able to follow parts of his argument, but I start to become unclear when/after he concludes that $\sup V_c < M$ or how continuity is really used in any of this. I have already seen this question regarding the same theorem, but unfortunately still unclear.
Best Answer
Let me write out case 2 in a bit more detail.
We look at two cases, $f(c) < M$ and $f(c) = M$. In the second case we are done, since $f$ assumes its maximum at $c$, so assume that $f(c) < M$. Then we have $M - f(c) > 0$, so we can choose an $\epsilon > 0$ such that $M - f(c) > \epsilon$. By continuity at $c$ (and it is here we use continuity!), there is a $\delta > 0$ such that $\lvert t - c \rvert < \delta$ implies $\lvert f(t) - f(c) \rvert < \epsilon$ for $t \in [a,b]$. This means that for all points $t \in (c - \delta, c]$ we have $f(t) < M - \epsilon$.
Since $c$ is the supremum of $X$, there must exist a point $t_0 \in X \cap (c - \delta, c]$, and by definition of $X$ we have $\sup V_{t_0} < M$. Then there must be some $\epsilon' > 0$ such that $f(t) < M - \epsilon'$ for all $t \in [a,t_0]$, by definition of $V_{t_0}$. Letting $\epsilon'' = \min\{\epsilon, \epsilon'\}$ we find that $f(t) < M - \epsilon''$ for all $t \in [a,c]$, which implies that $\sup V_c \leq M - \epsilon'' < M$ as desired.
Now if $c < b$, then there is some $t_1$ with $c < t_1 < b$, and such that $t_1 < c + \delta$. It follows that $f(t) < M - \epsilon$ for $t \in [c, t_1]$, and this was already true for $t \in [a,c]$, so it holds for all $t \in [a, t_1]$. But then $\sup V_{t_1} \leq M - \epsilon < M$, so $t_1 \in X$. But $c$ was the least upper bound of $X$, so this is a contradiction, and hence $c = b$. It follows that $\sup V_c = \sup V_b = M$, by definition of $M$. But this contradicts the assumption that $\sup V_c < M$ (that is, we have arrived at the inequality $M < M$), and so the assumption $f(c) < M$ must be false. Hence $f(c) = M$.
Notice that we only use continuity at $c$. We still need the assumption that $f$ is continuous on all of $[a,b]$, since we have no way of knowing where $c$ is!