Prove that $\lim \limits_{n\to+\infty}\|\sup_{k>n}u_k-u_n\|=0$ Let $\{u_n\}$ be a Cauchy sequence in the space $(\mathbb R,d)$ with $d(x,y)=\|x-y\|$.
Prove that $$\lim_{n\rightarrow+\infty}\|\sup_{k>n}u_k-u_n\|=0.$$
This seems to be obviously however I can not find an elegant proof.
 A: I am going to rewrite your statement to make it a little clearer. We need to show that
$$
\lim_{n \to \infty} \sup_{k > n} |u_k - u_n| = 0.
$$
Let me know if you object to these changes. There are several ways to go about proving this claim, and I am going to choose to do so by contradiction. Many people find proofs by contradiction to be distasteful if a direct proof is available, but I like them because I find it's often natural to proceed by asking "what happens if this statement isn't true?". In our case, this would mean that either the limit does not exist, or that it is greater than zero.
So suppose first that
$$
\lim_{n \to \infty} \sup_{k > n} |u_k - u_n| = \alpha > 0.
$$
We know that $\{u_k\}$ is Cauchy, and so we can find an integer $j$ so that $|u_m - u_k| < \alpha$ for all $m, k > j$. In particular, $|u_{j+1}, u_k| < \alpha$ for all $k > j+1$. Thus
$$
\sup_{k > j+1}|u_k - u_{j+1}| < \alpha.
$$
Now define $n = j+1$ so we can write this more simply:
$$
\sup_{k > n}|u_k - u_{n}| < \alpha.
$$
Note that this is true for all $n$ large enough, a contradiction (why?). Having seen the contradiction, you can go back and rewrite this more cleanly, even creating a direct proof if you prefer.
There is one consideration left: what if the limit does not exist? You can arrive at a contradiction here by following an analysis similar to the above. 
A: Let $\epsilon>0$ be given.  Let us prove that the absolute value is eventually at most $\epsilon$; if we can do that for any $\epsilon>0$, then the limit must be 0.
Because $(u_n)$ is Cauchy, there exists $N\in\mathbb{N}$ such that $\lvert u_n-u_m\rvert<\frac{\epsilon}{2}$ for all $n,m\geq N$.
In particular, this means $u_{N+1}\leq\sup_{k>N} u_k\leq u_{N+1}+\frac{\epsilon}{2}$. So, for $n>N$,
$$
\left\lvert \sup_{k>n} u_k-u_n\right\rvert\leq\Big\lvert\sup_{k>n}u_k-u_{N+1}\Big\rvert+\Big\lvert u_{N+1}-u_n\Big\rvert\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,
$$
so that the limit must also be at most $\epsilon$. This completes the proof.
