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I need help with this problem.

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

Thank you

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Do you know that every Cauchy sequence converges in the reals? Can a Cauchy sequence in $[0,1]$ converge outside $[0,1]$? –  Andrew Salmon Apr 4 '13 at 3:55
sorry you are wrong not every Cauchy sequence converge –  leena adam Apr 4 '13 at 4:01
Actually.... he's right. –  Euler....IS_ALIVE Apr 4 '13 at 4:02
The theorem didn't say that the converse is not true, did it? –  Ittay Weiss Apr 4 '13 at 4:07
@leenaadam in $\mathbb R$, every Cauchy sequence converges. –  Ittay Weiss Apr 4 '13 at 4:16

2 Answers 2

up vote 2 down vote accepted

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 $[0,1]$.

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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 $a\in [0,1]$.

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