Prove that if $(f_n)_{n \geq 1} \to f$ uniformly then $f_n \to f \in B(X)$

Question

Let $B(X)$ be the set of bounded functions from $X$ to $\mathbb{C}$ where $X$ is any metric space. Let $(f_n)_{n \geq 1} $ be a sequence in $B(X)$. Show that if $f_n \to f$ uniformly where $f:X\to \mathbb{C}$ is bounded then $f_n \to f$ in $B(X,d_{\infty})$.

Now, since $f_n \to f$ uniformly, given $\epsilon >0$ there exist $n_0 \in \mathbb{N}$ such that $|f_n(x)-f(x)|<\epsilon$ if $n\geq n_0$ for any $x\in X$. I wasn't sure, I could automatically imply that $sup_{x\in X}|f_n(x)-f(x)| < \epsilon$ since this is not true in general. However when reading my textbook, the author states that since $|f_n(x)-f(x)|<\epsilon, \forall x$ then $sup_{x\in X}|f_n(x)-f(x)| \leq \epsilon$. I assume he is using in such a way that $f_n$ is bounded. Is that step trivial?

Answer 1

The subtlety in the order of the quantifiers. Uniformity garuntees that the bound $$ |f_n(x)-f(x)|<\varepsilon $$ holds for all $x$ and all $n>n_0$ for the same $\varepsilon$. So, the $\sup |f_n(x)-f(x)| $ can't be greater than $\varepsilon$ if for each $x$ and $n>n_0$ the quantity $|f_n(x)-f(x)|$ isn't greater than $\varepsilon$.

Answer 2

There exists $n_0$ such that

$$|f_{n_0}(x)-f(x)|< 1, x \in X.$$

We have $|f_{n_0}(x)|< M$ on $X$ for some $M>0.$ It follows that

$$|f(x)|\le |f(x)-f_{n_0}(x)| + |f_{n_0}(x)| < 1 + M$$

for all $x\in X.$