# Why is the space of bounded functions complete?

Let $$(X,d_X),(Y, d_Y)$$ be two metric spaces. We define:

$$\begin{equation} B(X,Y)=\{f:X\rightarrow Y, f\text{ bounded}\} \end{equation}$$ Introducing the metric (of uniform convergence): $$\begin{equation} d(f,g)=\underset{x\in X}{\sup}d_Y(f(x),g(x)) \qquad\forall f,g\in B(X,Y) \end{equation}$$

Now, if $$Y$$ is complete, you can prove that $$B(X,Y)$$ is complete. I found this proof on a book, but there's a part I don't understand.

Proof

Let $$\{f_n\}_{n\geqslant1}$$ be a Cauchy sequence in $$B(X,Y)$$ and $$x\in X$$. The sequence $$\{f_n(x)\}$$ is a sequence in $$Y$$. $$\{f_n(x)\}$$ is trivially (so I'll be omitting some steps about this) a Cauchy sequence by definition of $$d$$: $$\begin{equation} d_Y(f_n(x),f_m(x))\leqslant\underset{x\in X}{\sup}d_Y(f_n(x),f_m(x))=d(f_n,f_m) \end{equation}$$

Since $$Y$$ is complete, $$\{f_n(x)\}$$ is convergent $$\begin{equation} \lim_{n\to+\infty} f_n(x)=f(x) \tag{1} \end{equation}$$

Let's write down the definition of Cauchy sequence for $$\{f_n\}$$: $$\begin{equation} \forall\epsilon>0 \quad \exists\nu\in\mathbb{N}: \forall n,m>\nu\quad n,m\in\mathbb{N}\implies d(f_n,f_m)<\epsilon \end{equation}$$ Thus, we have: $$\begin{equation} \forall x\in X\quad\forall\epsilon>0 \quad \exists\nu\in\mathbb{N}: \forall n,m>\nu\quad n,m\in\mathbb{N}\implies d_Y(f_n(x),f_m(x))<\epsilon \tag{2} \end{equation}$$

Well, this is the critical point. The book now takes the limit as $$m\to+\infty$$ and using $$(1)$$:

$$\begin{equation} d_Y(f_n(x), f(x))\leqslant\epsilon \quad \forall x\in X \quad \forall n>\nu \tag{3} \end{equation}$$

Since this is valid $$\forall x\in X$$, it follows that $$d(f_n, f)\leqslant\epsilon$$ and the theorem is proved.

As I said, I don't understand the step between $$(2)$$ and $$(3)$$. It looks like we are assuming $$d$$ is continuos when we take the limit. My guess is that he used the reverse triangle inequality and then took the limit but I'm not sure about that. So how do we get from $$(2)$$ to $$(3)$$?

Yes, they used that $$d$$ is continuous. In a metric space, the metric itself is always continuous.