Question in real analysis I need help with this problem.

Show that a Cauchy sequence in $[0,1]$ must converge to a point of $[0,1]$.

Thank you 
 A: In the real numbers every Cauchy sequence converges to some real number (this property is essentially equivalent to the completeness of the real numbers). A proof sketch is: A Cauchy sequence is bounded, thus, by the Bolzano-Weierstrass theorem, it has a convergent subsequence (this is the completeness bit actually). For a Cauchy sequence, the limit of a convergent subsequence must actually be the limit of the entire sequence. 
Now, you probably proved that weak limits are preserved by when passing to the limit. That is, if a sequence $a_n$ converges to $a$, and $a_n\le t$, then $a\le t$. Similarly for inequalities in the other direction. So now, you have a sequence in $[0,1]$ which is Cauchy. So it converges to some point. But by preservation of weak inequalities, the limit point $a$ satisfies $a\in [0,1]$.
A: Suppose a Cauchy sequence converges outside $[0,1]$.  So $x_n$ is Cauchy in $[0,1]$ and converges to $x \notin [0,1]$.  If $x > 1$, then take $x - 1 > 0$, and if $x < 0$, take $-x >0$.
Take $0 < \epsilon < -x$ or $0 < \epsilon < x-1$ (depending on the cases above).  Since $x_n \to x$, then $|x_n - x| < \epsilon$ for $n \ge N$ for some $N$.  If $x > 1$, then $|x_n - x| \ge x - 1 > \epsilon$, which is a contradiction.  If $x < 0$, then $|x_n - x| \ge -x > \epsilon$, which is a contradiction.
But since $x_n \to x$, for some $x$ (since every real Cauchy sequence converges), $x$ must be in $[0,1]$.
