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Let $S\subset M$ dense subset. Given a sequence $(x_n)$ in $M$, suppose that for some $x\in M$ $\lim d(x_n,s) = d(x,s)$ for every $s\in S$. Prove that $\lim x_n = x$.

My attempt: Since $S$ is dense in $M$ and $x\in M$, it follows that $x = \lim a_n$ for some sequence $(a_n)$ in $S$. It is given that $\lim d(x_n,s) = d(x,s)$ for every $s\in S$. So, in particular we can evaluate for all elements of $(a_n)$. Therefore, noting that the metric function $d$ is continuous, we obtain $\lim d(x_n,a_n) = d(x,a_n) = d(\lim a_n,a_n) = \lim d(a_n,a_n) = 0.$

But this only proves that $\lim d(x_n,a_n) = 0$. How can I proceed to the conclusion that $\lim x_n = x$?

John Griffin
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user2345678
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3 Answers3

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hint

$$\lim_{n\to+\infty}d (x_n,a_n)=0$$

means that if $(a_n) $ converges, $(x_n) $ will also converge and to the same limit, since we have

$$0\le d (x_n,x)\le d (x_n,a_n)+d (a_n,x) $$

hamam_Abdallah
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What you have written isn't quite correct because you have two sequences indexed by $n$, but you are only taking the limit of one of them. Every $k$ satisfies $$ \lim_{n\to\infty}d(x_n,a_k) = d(x,a_k), $$ so by the triangle inequality, we have $$ \limsup_{n\to\infty} d(x_n,x) \le \limsup_{n\to\infty}d(x_n,a_k) + d(a_k,x) = 2d(x,a_k). $$ Since this holds for every $k$ and $\lim_{k\to\infty}d(x,a_k)=0$, we conclude $$ \lim_{n\to\infty}d(x_n,x)=0. $$

John Griffin
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  • Why did you chose to evaluate the $\lim\sup$? – user2345678 Oct 15 '17 at 18:03
  • @math.h At that moment in the proof, it is not known whether $\lim_{n\to\infty}d(x_n,x)$ exists. However the limit supremum always exists, and if I can show that it must be zero, then (by the fact that distances are positive), we know that the limit exists and is zero. – John Griffin Oct 15 '17 at 18:04
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    Perfect care, thank you. – user2345678 Oct 15 '17 at 18:06
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for all $s\in S$

$$\lim d(x_n,x) \le d(x,s)+\lim d(x_n,s) = 2d(x,s)$$

But by density there exists $s_n\in S$ with $$\lim s_n, =x$$ that is $$ \lim d(s_n,x) = 0 $$ so in particular for $s_n$, $$\lim d(x_n,x) \le 2d(x,s_n)$$ for all n . letting n to $\to\infty$ we get
$$\lim d(x_n,x) \le\lim 2d(x,s_n) =0$$ that is $$\lim d(x_n,x)=0$$

Guy Fsone
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