Complete unbounded interval Why is $[1,\infty)$ complete? 
I know that every Cauchy sequence should converge for the space to be complete, but I cannot see it.
Taking a sequence in this space then $\forall \epsilon>0, \exists  n>0$ s.t. if m>n then $d(x_n,x_m)<\epsilon$
A sequence in this space converges iff $\forall \epsilon>0, \exists  n>0$ s.t. if m>n then $d(x_m,l)<\epsilon$
 A: If we have a Cauchy sequence in $A = [1,\infty)$, so $(x_n)$ (all in $A$) with the property that $\forall \epsilon>0: \exists k: \forall m,m \ge k: d(x_n, x_m) < \epsilon$, where $d$ denotes the Euclidean metric, then this sequence is also a Cauchy sequence in $\mathbb{R}$ (using the same metric $d$ of course): we use nothing about points being in $A$ or not: the property is intrinsic to the sequence and the used metric.
But as $\mathbb{R}$ is complete there is some $x \in \mathbb{R}$ such that $x_n \to x$ as $n \to \infty$.
If $x \notin A$, then there is some ball $B(x,r), r>0$ that misses $A$, because $A$ is a closed set. But then there can be no points of $A$ within distance $r$ of $x$, and this contradicts $x_n \to x$. So $x \in A$. But now $x_n \to x$ in $A$ (because the limit is in $A$ and we use the same distance). So we have shown that all Cauchy sequences in $A$ converge to a point in $A$, just using that the ambient space ($\mathbb{R}$ in our case) is complete and the subspace $A$ is closed. It is essential we are using the same metric on both.
