= Solution
{c}
The <sequence> $(f_n)$ converges by <uniform convergence> to $f$ on $E$ if
$$
\forall\varepsilon>0\ \exists M\in\mathbb N\ \forall n\geq M\ \forall x\in E:\quad |f_n(x)-f(x)|<\varepsilon.
$$
The index $M$ must work simultaneously for every $x\in E$.
Given $\varepsilon>0$, choose $m\geq N$ so large that $|f_m(x)-f(x)|<\varepsilon/3$ for every $x\in E$. The <continuity> of this one fixed $f_m$ at $x_0$ gives $\delta>0$ such that $x\in E$ and $|x-x_0|<\delta$ imply $|f_m(x)-f_m(x_0)|<\varepsilon/3$. The <triangle inequality> then gives
$$
|f(x)-f(x_0)|\leq|f(x)-f_m(x)|+|f_m(x)-f_m(x_0)|+|f_m(x_0)-f(x_0)|<\varepsilon.
$$
Thus \b[$f$ is continuous at $x_0$], with <continuity> understood relative to $E$. This is the pointwise form of the <uniform limit theorem>.
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