Normed space of bounded functions $f:\mathbb{N}\to\mathbb{N}$ Let $X = \{f:\mathbb{N}\to\mathbb{N}: \exists M\in\mathbb{N} \forall n\in\mathbb{N} f(n) \leq M\}$.
Define a norm on $X$ by defining for $f\in X$: $$||f|| = \sum_{n=1}^\infty \frac{f(n)}{2^n}.$$
Is the metric space associated $(X,||\cdot ||)$ complete? What about compactness?
 A: Well this is not really a norm (because $X$ is not even an abelian group) nevertheless $X$ can be always seen as a metric space with the following distance :
$$d(f,g):=\sum_{n=1}^{\infty}\frac{|f(n)-g(n)|}{2^n} $$
I think now of $X$ with the distance written above.  Define a sequence $(f_k)$ :
$$f_k(n):=n\text{ if } n\leq k\text{ and } 0 \text{ if } n>k $$
Then $f_k\in X$ (it is obvious) furthermore, if $k\leq l$ then :
$$d(f_k,f_l)=\sum_{n=k+1}^l\frac{n}{2^n}\leq \sum_{n=k+1}^{\infty}\frac{n}{2^n} $$
The last part converging toward $0$ when $k$ goes to infinity independently of $l$. This shows that $(f_k)$ is a Cauchy sequence. 
Now, here is an aside affirmation that is easily verified :

If $(g_k)$ converges in $X$ toward $g$ then $g_k(n)\rightarrow_k g(n)$ in $\mathbb{N}$ for any $n$.

Now, assume that our sequence $(f_k)$ had a limit $f$ then, by what is written above : $f(n)=lim_kf_k(n)$. It is clear that $lim_kf_k(n)=n$ so that the limit function sends any integer $n$ to itself. But this function is not bounded hence it is not in $X$, this shows that the Cauchy sequence $(f_k)$ does not converge hence $X$ is not complete and in particular not compact.
