Limit of a sequence on a metric space

[mahler1]

Homework Statement .

Let $(X,d)$ be a metric space and let $D \subset X$ a dense subset of $X$. Suppose that given $\{x_n\}_{n \in \mathbb N} \subset X$ there is $x \in X$ such that $\lim_{n \to \infty}d(x_n,s)=d(x,s)$ for every $s \in D$. Prove that $\lim_{n \to \infty} x_n=x$.

The attempt at a solution.

What I did was:
Let $\epsilon>0$
$d(x_n,x)\leq d(x_n,s)+d(s,x)$. For each $n \in \mathbb N$ I know there is $s \in D$ such that $d(x_n,s)<\dfrac{\epsilon}{2}$. But, this doesn't say anything, I don't know how find the $N \in \mathbb N$ such that for all $n\geq N$, $d(x_n,s)+d(s,x)\leq \epsilon$

 

[PeroK]

You know that D is dense in X. So, either:

x $\in$ D (in which case the result follows)

or

There is a sequence in D converging to x.

Can you take it from there?

It's a bit tricky, this one!

 

[mahler1]

You know that D is dense in X. So, either:

x $\in$ D (in which case the result follows)

or

There is a sequence in D converging to x.

Can you take it from there?

It's a bit tricky, this one!

I'll give it another try with your suggestion:

As $D$ is dense in $X$, there is a sequence $\{s_n\}_{n \in \mathbb N}$ such that $s_n \to x$ when $n \to \infty$.

By the triangle inequality,
$0\leq d(x_n,x)\leq d(x_n,s_n)+d(s_n, x)$. Again I have problems, because now I know that for a given $\epsilon$, I can pick $n_0$ : for all $n\geq n_0 \space$ $d(s_n,x)<\dfrac{\epsilon}{2}$, but how can I control the distance $d(x_n,s_n)$?

 

[PeroK]

how can I control the distance $d(x_n,s_n)$?

You know something about $d(x_n,s_m)$ for any m. By choosing the same subscript for the two sequences, you boxed yourself in a bit!

 

[mahler1]

You know something about $d(x_n,s_m)$ for any m. By choosing the same subscript for the two sequences, you boxed yourself in a bit!

Sure, thanks for the help.

First, consider $n_0$ : $d(s_n,x)<\dfrac{\epsilon}{4}$ for all $n\geq n_0$.

Now, I know that $d(x_n,s) \to d(x,s)$ for every $s \in D$. Then, there exists $n_1$ such that for all $n\geq n_1$, $|(d(x_n,s)-d(x,s)|<\dfrac{\epsilon}{2}$ for all $s \in D$. This means $d(x_n,s)<\dfrac{\epsilon}{2}+d(x,s)$ for all $s \in D$ and for all $n\geq n_1$.

If this inequality holds for every element in $D$, in particular, it holds for every $s_n \in \{s_n\}_{n \in \mathbb N}$. If I choose $N=\max\{n_0,n_1\}$, then, for all $n\geq N$, $0\leq d(x_n,s)\leq d(x_n,s_n)+d(s_n,x)<2d(s_n,x)+\dfrac{\epsilon}{2}<2\dfrac{\epsilon}{4}+\dfrac{\epsilon}{2}=\epsilon$.

 

[PeroK]

You've got the right idea, but you definitely need to use different subscripts for the two sequences. You've got a moving target for the convergence of d(x_n, s_n). Do you see the slight problem with what you've got?

It's also a good idea in general, because when you have two sequence like this, you don't always want to compare x_n with s_n. You want to fix a given member of s: s_m, say, and then consider the sequence d(x_n, s_m) as n →∞