[proofplan] [Uniform convergence](/page/Uniform%20Convergence) gives one index threshold that works simultaneously for every point of $S$. To prove pointwise convergence, we fix a single point $s\in S$ and specialize the uniform estimate to that point. Since the point was arbitrary, the metric convergence holds at every point of $S$. [/proofplan] [step:Specialize the uniform estimate to an arbitrary point] If $S=\varnothing$, then the assertion “for every $s\in S$” is vacuous, so the conclusion follows. Assume now that $S\neq\varnothing$, and fix an arbitrary point $s\in S$. Let $\varepsilon>0$ be given. Since $f_k$ converges uniformly to $f$ on $S$, there exists $N\in\mathbb N$ such that for every $k\in\mathbb N$ with $k\ge N$ and every $x\in S$, \begin{align*} e(f_k(x),f(x))<\varepsilon. \end{align*} Taking $x=s$ in this estimate is allowed because $s\in S$. Hence, for every $k\ge N$, \begin{align*} e(f_k(s),f(s))<\varepsilon. \end{align*} This is exactly the $\varepsilon$-$N$ condition for the sequence $(f_k(s))_{k\in\mathbb N}$ to converge to $f(s)$ in the [metric space](/page/Metric%20Space) $(Y,e)$. [guided] We want to prove pointwise convergence, which means that after fixing a point $s\in S$, the sequence of points $(f_k(s))_{k\in\mathbb N}$ in the metric space $(Y,e)$ must converge to $f(s)$. If $S=\varnothing$, there is no point $s$ to check, so the universal conclusion is vacuous. Thus suppose $S\neq\varnothing$, and fix an arbitrary point $s\in S$. Let $\varepsilon>0$ be given. Uniform convergence of $f_k$ to $f$ on $S$ means that this $\varepsilon$ has a single index threshold $N\in\mathbb N$ which works for all points of $S$ at once. Therefore there exists $N\in\mathbb N$ such that whenever $k\ge N$ and $x\in S$, we have \begin{align*} e(f_k(x),f(x))<\varepsilon. \end{align*} Now we use the only difference between uniform and pointwise convergence. Pointwise convergence at $s$ only asks for the estimate at this one fixed point. Since the uniform estimate holds for every $x\in S$, and since our chosen point satisfies $s\in S$, we may substitute $x=s$. This gives, for every $k\ge N$, \begin{align*} e(f_k(s),f(s))<\varepsilon. \end{align*} The number $N$ may depend on $\varepsilon$, but it does not need to depend on anything else now that $s$ has been fixed. Thus the sequence $(f_k(s))_{k\in\mathbb N}$ converges to $f(s)$ in the metric space $(Y,e)$. [/guided] [/step] [step:Conclude pointwise convergence on the whole set] The point $s\in S$ was arbitrary. Therefore, for every $s\in S$, the sequence $(f_k(s))_{k\in\mathbb N}$ converges to $f(s)$ in $(Y,e)$. Hence $f_k$ converges pointwise to $f$ on $S$. [/step]