# $x_n \rightarrow x \implies \lim_{n \rightarrow +\infty}$ IFF $f_n \rightarrow f$ uniformly in each $K \subset M$ compact

Let $f, f_1, f_2, \dots$ be continuous maps from M to N, M and N metric spaces. The, the following affirmations are equivalent:

$(a)$$x_n \rightarrow x \implies \lim_{n \rightarrow +\infty} f_n(x_n) = f(x) (b)$$ f_n \rightarrow f$ uniformly in each $K \subset M$ compact

• $(a) \implies (b)$ Fix $K \subset M$ compact. We have to prove that given $\epsilon > 0$, there exists $n_0$ such that $n>n_0 \implies d(f_n(x), f(x)) < \epsilon$, for each $x \in K$.

As $\lim f_n(x_n) = f(x)$, there exists $n_o$ such that $n>n_o \implies d(f_n(x_n), f(x)) < \epsilon /4$. As K is compact, $f_{n_o}$ is uniformly continuous, that is, there exists $\delta > 0$ such that $d(x, y) < \delta \implies d(f_{n_0}(x), f_{n_0}(y)) < \epsilon / 4$.

I think I should not put in this order, because I am trying to do some triangle inequalities and I can't conclude...

• $(b) \implies (a)$

As $f_n \rightarrow f$ uniformly, given $\epsilon > 0$, there exists $N$ such that $n> N \implies d(f_n(x), f(x)) < \epsilon/2.$

Moreover, as $x_n \rightarrow x$, there is $M$ such that $n > M \implies d(x_n, x) < \delta$, and, as $f$ is continuous, this implies that $d(f(x_n), f(x)) < \epsilon/2$

Then, if $n> max\{N, M\}$:

$d(f_n(x_n), f(x)) \leq d(f_n(x_n), f(x_n)) + d(f(x_n), f(x)) < \epsilon/2 + \epsilon /2$.

$(a) \implies (b)$ Fix $K \subset M$ compact. We have to prove that given $\epsilon > 0$, there exists $n_0$ such that $n>n_0 \implies d(f_n(x), f(x)) < \epsilon$, for each $x \in K$.
Assume the converse. That is for each $n$ there exists a point $x_n\in K$ such that $d(f_n(x), f(x)) \ge \epsilon$. Since the set $K$ is a metric compact, a sequence $\{x_n\}$ has a subsequence $\{x_{n_k}\}$ convergent to a point $x\in X$. Condition (a) imply that a sequence $\{f_{n_k}(x_{n_k})\}$ converges to $f(x)$. The continuity of the map $f$ implies that a sequence $\{f(x_{n_k})(x)\}$ converges to $f(x)$ too. But $d(f_{n_k}(x_{n_k}), f(x_{n_k}))\ge\epsilon$ for each $n_k$, a contradiction.
$(b) \implies (a)$ Since a sequence $\{x_n\}$ converges to the point $x$, the set $K=\{x\}\cup\{x_n\}$ is compact. Let $\epsilon>0$ be an arbitrary. As $f_n \rightarrow f$ uniformly on the set $K$, there exists a number $n_1$ such that $d(f_n(y), f(y)) < \epsilon/2$ for each $y\in K$ and each $n>n_1$. The continuity of the map $f$ implies that there exists a number $n_2$ such that $d(f_n(x), f(x)) < \epsilon/2$ for each $n>n_2$. Then for each $n>\max\{n_1, n_2\}$ we have $$d(f_n(x_n), f(x)) \leq d(f_n(x_n), f(x_n)) + d(f(x_n), f(x)) < \epsilon/2 + \epsilon /2=\epsilon.$$