Show that $(f_n)$ converges uniformly to $f$ .

Question

Let $(X, d)$ be a compact metric space.

Suppose $f$ and $(f_n)$ are real-valued continuous functions on $X$.

Suppose that, for each $x\in X$, the sequence $(f_n(x))$ is a monotonic sequence converging to $f(x)$.

Show that $(f_n)$ converges uniformly to $f$.

(Hint: Given $\epsilon > 0$, show that the sets $U_n = \{x \in X : | f_n(x) − f (x)| < \epsilon\}, n \in \mathbb{N}$, form an open cover of $X$)

I do not really understand what this hint can do to solve the problem. Thank you.

Answer 1

Let $\epsilon > 0$.

Once you have $U_n = \{x \in X : | f_n(x) − f (x)| < \epsilon\}, n \in \mathbb{N}$, form an open cover of $X$, since $X$ is compact, there is an $N \in \mathbb{N}$ such that $X \subset \bigcup_{i=1}^{N}U_i$. Since $U_i \subset U_j$ for $i \leq j$, then $\bigcup_{i=1}^{N}U_i=U_N$, and hence $X \subset U_N \subset U_{N+1} \cdots$

Therefore, for all $x \in X$ we have $|f_n(x)-f(x)|< \epsilon$ for all $n \geq N.$

Answer 2

Well, since $X$ is compact you only need a finit enumber of elements to cover the whole set. Since $U_n \subseteq U_m$ for $n\leq m$ we have that there is an $N$ such that $U_N,U_{n+1},\dots $ are all equal to $X$, which is what you wanted to prove.