Max and Min of Function on Closed Real Interval/Proof 1

Proof
From Image of Closed Real Interval is Bounded, we have that $f$ is bounded on $\closedint a b$.

Let $d$ be the supremum of $f$ on $\closedint a b$.

Consider a sequence $\sequence {x_n}$ in $\closedint a b$ such that $\size {\map f {x_n} } \to d$ as $n \to \infty$.

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

Now $\closedint a b$ is a closed interval

So from Convergent Subsequence in Closed Interval, $\sequence {x_n}$ has a subsequence $\sequence {x_{n_r} }$ which converges to some $\xi \in \closedint a b$.

Because $f$ is continuous on $\closedint a b$, it follows from Limit of Image of Sequence that $\map f {x_{n_r} } \to \map f \xi$ as $r \to \infty$.

So $\map f \xi = d$ and thus the supremum $d$ is indeed a maximum.

A similar argument shows that the infimum is a minimum.