For any
\(\epsilon > 0\text{,}\) let
\(\delta > 0\) be such that
(2.6.1) holds. Since
\(K\) is compact, it can be covered by a finite number of sets of diameter
\(< \delta\text{,}\) say,
\(V_{1}, \cdots, V_{m}\text{.}\)
Pick \(p_{i} \in V_{i}\) for each \(i=1,\cdots, m\text{.}\) Then for each \(i=1,\cdots,m\text{,}\) there exists some \(M_{i} > 0\) such that \(|f_{n}(p_{i})|\le M_{i}\) for all \(n\text{.}\) Any \(x\in V_{i}\) satisfies
\begin{equation*}
|f_{n}(x)-f_{n}(p_{i})| < \epsilon, \text{ therefore } |f_{n}(x)|\le M_{i}+\epsilon.
\end{equation*}
Let \(M=\max_{1\le i\le m}M_{i}\text{.}\) Then any \(x\in K\) satisfies \(|f_{n}(x)|\le M+\epsilon\) for all \(n\text{,}\) which shows (a).
The compactness of
\(K\) implies that it has a dense countable subset
\(\{q_{i}\}\text{.}\) It then follows from
Proposition 2.6.4 that we can pick a subsequence
\(\{f_{n_{k}}\}\) such that
\(\{f_{n_{k}}(q_{i})\}\) converges for each
\(q_{i}\text{,}\) \(i=1,\cdots\text{.}\)
Each
\(V_{i}\) contains some
\(q_{j_{i}}\text{.}\) It follows that there exists some
\(N\) such that
\(|f_{n_{k}}(q_{j_i})-f_{n_{l}}(q_{j_i})| < \epsilon\) for all
\(k, l\ge N\) and
\(i=1,\cdots, m\text{.}\)
Using the finite cover
\(\{V_{i}\}_{i=1}^{m}\) of
\(K\) and
(2.6.1) on
\(\{f_{n_{k}}\}\text{,}\) we find that for any
\(x\in K\text{,}\) \(d(x, q_{j_i}) < \delta \) for some
\(i=1,\cdots, m\text{,}\) therefore,
\begin{align*}
\amp |f_{n_{k}}(x)-f_{n_{l}}(x)|\\
\le \amp |f_{n_{k}}(x)-f_{n_{k}}(q_{j_i})|+| f_{n_{k}}(q_{j_i})-f_{n_{l}}(q_{j_i})|
+ |f_{n_{l}}(q_{j_i})-f_{n_{l}}(x)| \le 3\epsilon
\end{align*}
for all \(k, l\ge N\text{.}\) This show that \(\{f_{n_{k}}\}\) is uniformly Cauchy on \(K\text{,}\) therefore proving (b).
