# showing a space is complete

Let $X$ be the set of real bounded functions. Let $(X,\|f-g\|_\sup)$ be a metric space. Prove that this space is complete.

My proof: Let $\{f_n\}$ be a cauchy sequence. So we have $$\exists N, \forall n,m \geq N, \|f_n-f_m\|_\sup = \sup_x |f_n - f_m| < \epsilon$$ But we have $|f_n - f_m| \leq \sup_x |f_n-f_m| < \epsilon$ and since this is true for every x we have $|f_n - f_m| < \epsilon, \forall x$ Hence, $\{f_n\}$ is a uniform cauchy sequence. Hence it converges uniformly.

Now I need to show that limit is in $X$. I'm not sure how to do this. Any help?

• I changed $\displaystyle||f_n-f_m||$ to $\displaystyle\|f_n-f_m\|$, which is standard. In at least some contexts the difference is conspicuous, and in most (or all?) it is clearly visible. ${}\qquad{}$ Mar 8 '15 at 18:05

Let $f$ be the pointwise limit of the sequence $(f_n)$, i.e. $f(x) = \lim_{n\to\infty} f_n(x)$ for all $x\in\mathbb{R}$. The limit is, of course, well-defined.
What's a good way to bound $|f(x)|$ uniformly? Well, the only information we've got about the size of anything is on $|f(x) - f_n(x)|$ (uniform convergence). So we'll try the triangle inequality:
$$|f(x)| \leq |f(x) - f_n(x)| + |f_n(x)|.$$
Is there a way to uniformly bound the left-hand side? (Hint: Where does each $f_n$ lie? Can you pick a good $n$ to make things uniformly bounded?)