My subset $X$ has the Bolzano-Weierstrass property and I need to prove that $X$ is complete in the sense that every Cauchy sequence in $X$ converges to a point in $X$.

I know that having the Bolzano-Weierstrass property means that every sequence $(x_n)_{n=1}^{\infty}\subset X$ has a convergent subsequence $x_{n_k} \xrightarrow{k\to\infty} x\in X$, but I'm unsure how to use this to prove $X$ is complete.

  • $\begingroup$ If a sequence is Cauchy, then it has at most one limit point. $\endgroup$ – egreg Nov 21 '14 at 9:47
  • $\begingroup$ I know that every Cauchy sequence in $X$ has a convergent subsequence that converges to a point in $X$. Does it follow from this that the Cauchy sequences all have one limit? And then how do I know that the limit is in $X$? $\endgroup$ – user184036 Nov 21 '14 at 9:50

Let $\varepsilon>0$; then you know that you can find $r$ such that:

  1. for every $m,n>r$, $\|x_m-x_n\|<\varepsilon/2$
  2. for every $k>r$, $\|x_{n_k}-x\|<\varepsilon/2$

So, suppose $n>r$; then, for some $k>r$, you have $n_k>n$ and $$ \|x_n-x\|\le\|x_n-x_{n_k}\|+\|x_{n_k}-x\|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2} $$

  • $\begingroup$ So then this means that $\| x_n - x \| < \epsilon $ and therefore $x_n$ converges to $x\in X$ which means every Cauchy sequence in $X$ converges to a point in $X$, hence $X$ is complete? $\endgroup$ – user184036 Nov 21 '14 at 9:56
  • $\begingroup$ @john.smith Yes. Intuitively, the terms of a Cauchy sequence are eventually near to each other; if a subsequence converges, then the sequence is eventually near to this point. $\endgroup$ – egreg Nov 21 '14 at 9:59

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