Max and Min of Function on Closed Real Interval

Theorem
Let $$f$$ be a real function which is continuous on the closed interval $$\left[{a \,. \, . \, b}\right]$$.

Then $$f$$ reaches a maximum of $$d$$ and a minimum of $$c$$ on $$\left[{a \,. \, . \, b}\right]$$.

Proof
From Image of Closed Real Interval is Bounded, we have that $$f$$ is bounded on $$\left[{a \,. \, . \, b}\right]$$.

Let $$d$$ be the supremum of $$f$$ on $$\left[{a \,. \, . \, b}\right]$$.

Consider a sequence $$\left \langle {x_n} \right \rangle$$ in $$\left[{a \,. \, . \, b}\right]$$ such that $$\left|{f \left({x_n}\right)}\right| \to d$$ as $$n \to \infty$$.

From the corollary to Limit of Sequence to Zero Distance Point, this can always be found.

Now $$\left[{a \,. \, . \, b}\right]$$ is a closed interval

So from Convergent Subsequence in Closed Interval, $$\left \langle {x_n} \right \rangle$$ has a subsequence $$\left \langle {x_{n_r}} \right \rangle$$ which converges to some $$\xi \in \left[{a \,. \, . \, b}\right]$$.

Because $$f$$ is continuous on $$\left[{a \,. \, . \, b}\right]$$, it follows from Limit of Image of Sequence that $$f \left({x_{n_r}}\right) \to f \left({\xi}\right)$$ as $$r \to \infty$$.

So $$f \left({\xi}\right) = d$$ and thus the supremum $$d$$ is indeed a maximum.

A similar argument shows that the infimum is a minimum.