Equicontinuity plus dense-set convergence implies uniform convergence on compacta

Lemma (equicontinuity + dense-set convergence): Let $K$ be a compact metric space and let $(f_n)$ be a sequence of real-valued continuous functions on $K$ that is equicontinuous : for every $\varepsilon>0$ there exists $\delta>0$ such that $d(x,y)<\delta \implies |f_n(x)-f_n(y)|<\varepsilon \quad \text{for all } n.$ Assume there exists a dense subset $D\subseteq K$ such that for every $x\in D$, the sequence $(f_n(x))$ converges (equivalently, is Cauchy ).

Then $(f_n)$ is uniformly Cauchy on $K$, hence converges uniformly on $K$ to some continuous function $f$. Moreover, $f(x)=\lim_{n\to\infty} f_n(x)$ for all $x\in D$.

This lemma is a standard compactness-and-equicontinuity upgrade principle and is one of the key steps behind Arzelà–Ascoli-type arguments.

Proof sketch: Fix $\varepsilon>0$ and choose $\delta>0$ from equicontinuity for $\varepsilon/3$. Compactness gives a finite $\delta$-net in $K$; by density, choose net points $y_1,\dots,y_N\in D$. Since $(f_n(y_i))$ converges for each $i$, it is Cauchy, so choose $N_0$ making $|f_n(y_i)-f_m(y_i)|<\varepsilon/3$ for all $i$ when $m,n\ge N_0$. For any $x\in K$, pick $i$ with $d(x,y_i)<\delta$ and use $|f_n(x)-f_m(x)| \le |f_n(x)-f_n(y_i)|+|f_n(y_i)-f_m(y_i)|+|f_m(y_i)-f_m(x)|<\varepsilon,$ giving uniform Cauchy.