Proving EVT with epsilon-delta

Question

Just wondering whether we can prove Extreme Value Theorem purely using $\epsilon$-$\delta$ definition of continuity? Can you show me that proof, if possible? Cheers!

Answer 1

A lemma to this theorem is that a function continuous on a closed, bounded interval is bounded on this interval. Assuming that you have proved this lemma:

Let $ f:[a,b] \to \mathbb{R}\ ,\ S = \{\,f(x) \ | \ x \in [a,b]\} $ and $ M := \sup S$

We need to find an $x_M \in [a,b]$ such that $f(x_M) = M$

It is a theorem that one can always choose a sequence of points in a set that converges to that set's supremum.

Thus we can choose $y_n \in S$ such that $y_n \to M$. By definition of $S$, $y_n = f(x_n)$ for some $x_n \in [a,b].$ Since $ a \leq x_n \leq b$ for all $n$, $\{x_n\}$ is bounded. By the Bolzano-Weirstrauss Theorem, $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$, Say $x_{n_k} \to x_M$. By the comparison theorem for sequences, $x_M \in [a,b]$.

Finally, using the sequential characterization of continuity, $$x_{n_k} \to x_M \implies f(x_{n_k}) \to f(x_M)$$

However, $f(x_{n_k}) = y_{n_k}$ is a subsequence of $y_n$ and hence must converge to the same limit as $y_n.$

Thus we also have $f(x_{n_k}) \to M$.

By uniqueness of limits we must then have $f(x_M) = M$ and we're done. The case for infimums is virtually identical.