Exercise 7.9

The problem didn’t state it explicitly, but in order to apply the Chapter’s Theorems let’s assume that $E$ is a set in a metric space.

Let $𝜀>0$. Since $\{f_n\}$ converges uniformly to $f$, there is an integer $N_1$ such that if $n\ge N_1$ then for any $y\in E$ we have

Since $f$ is continuous by Theorem 7.12, there is an integer $N_2$ such that if $n\ge N_2$ then

Hence, if $n\ge \max <!--\; FUNCTION\; APPLICATION\; -->(N_1,N_2)$, we have

The converse is not true. For example, let $E=[0,1)$ and let $f_n(x)=x^n$. Then this sequence converges to the constant function $f(x)=0$, but not uniformly. If $\{x_n\}$ is a sequence of points of $E$ converging to a point $x$ in $E$, then it is contained in a closed subinterval $[0,a]$ where $a<1$. Since $f_n$ converges uniformly on this subinterval, $f_n(x_n)$ converges to $f(x)$ by the first part of this problem. Hence this property is insufficient for determining that the convergence is uniform.