Theorem 7.3.5 (Dini).label Let $X$ be a compact topological space, $\angles{f_\alpha}_{\alpha \in A}\subset C(X; \real)$, and $f \in C(X; \real)$ such that $f_{\alpha} \upto f$ pointwise, then $f_{\alpha} \upto f$ uniformly.
Proof. Let $\eps > 0$, then for each $\alpha \in A$, $\bracs{f - f_\alpha < \eps}\subset X$ is open, and $X = \bigcup_{\alpha \in A}\bracs{f - f_\alpha < \eps}$. By compactness, there exists $B \subset A$ finite such that $X = \bigcup_{\beta \in B}\bracs{f - f_\beta < \eps}$. Let $\alpha_{0} \in A$ with $\alpha_{0} \ge \beta$ for all $\beta \in B$, then since $f_{\alpha} \upto f$ pointwise,
\[\bracs{f - f_\alpha < \eps}= \bigcup_{\beta \in B}\bracs{f - f_\beta < \eps}= X\]
for any $\alpha \ge \alpha_{0}$.$\square$
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