6

I am having trouble proving the following statement:

"Let $(X,d)$ be a metric space and $A$ a dense subset such that every cauchy sequence in $A$ converges to a point in $X$. Prove that $X$ is complete".

Let us take a cauchy sequence $\{x_i\}$ containing points from both $A$ and $X\setminus A$.

If this cauchy sequence has a limit point, then it has to exist within $X$. Proof: Let us assume the limit $l$ exists outside $X$. Then every open set $B(l,r)$, $\forall r\in \Bbb{R}$ contains points from $\{x_i\}$, and because every open set containing any points from $X\setminus A$ contains points from $A$, every such $B(l.r)$ also contains points from $A$. Hence, $l$ should have been included in $\overline{A}=X$.

However, I can't prove that such a limit point exists in the first place. Any help would be greatly appreciated.

Thanks!

hudialala
  • 61
  • 2
    Given $(x_i)$ a Cauchy sequence in $X$, for each $i$ choose $a_i\in A$ with $d(a_i,x_i)<1/2^i$. Show $(a_i)$ is Cauchy. So $(a_i)$ converges to some $a$. Now show $(x_i)$ converges to $a$ as well. – David Mitra Aug 12 '13 at 16:12
  • What do you mean by "outside of $X$" ? – Stefan Hamcke Aug 12 '13 at 16:13
  • Another way to think about it: $X$ is by definition the completion of $A$, and the completion of a metric space is complete – BBBB Aug 12 '13 at 16:23

1 Answers1

9

A limit point can't "exist outside the space". You need to show that an arbitrary Cauchy sequence in $X$ converges. To do this, the basic idea is that given a Cauchy sequence $(x_i)$ from $X$, you can consider a sequence from $A$ that is "very close" to $(x_i)$.

Here is an outline of how to effect this:

  1. Given $(x_i)$ a Cauchy sequence in $X$, for each $i>0$ choose $a_i\in A$ with $d(a_i,x_i)<1/i$.

  2. Using the triangle inequality, show that $(a_i)$ is Cauchy (note here that you can write $$a_i-a_j=(a_i-x_i)+(x_i-x_j)+(x_j-a_j).$$

  3. So $(a_i)$ converges to some $a\in X$.

  4. Now, using the triangle inequality again, show that $(x_i)$ converges to $a$ as well.

David Mitra
  • 76,313
  • 1
    To prove that $(a_i)$ is cauchy, we have to prove that for any $\epsilon$, there is an $N$ such that $\forall m,n>N$, $d(a_m,a_n)<\epsilon$. $d(a_m,a_n)\leq d(a_m,x_m)+d(x_m,x_n)+d(x_n,a_n)<\frac{1}{m}+\frac{1}{n}+d(x_m,x_n)$. As per the conditions, $d(x_m,x_n)<\epsilon- \frac{1}{m}-\frac{1}{n}$. If we can find an $N_0$ such that $d(x_m,x_n)<\epsilon-\frac{2}{N_0}\forall m,n>N_0$, $N=N_0$. Is this correct? –  Aug 12 '13 at 17:59
  • 2
    @AyushKhaitan Given $\epsilon>0$, I would first choose $N$ with $1/N<\epsilon/3$. Then choose $N_0\ge N$ so that for $n,m\ge N_0$, one has $d(x_n,x_m)<\epsilon/3$. – David Mitra Aug 12 '13 at 18:52