# In a compact metric space, a sequence with a certain property is convergent

Mathematics Asked by StAKmod on December 19, 2020

The question:

Let $$(K,d)$$ be a compact metric space, if $${x_n}$$ is a sequence such that $${d(x_n,x)}$$ converges for all $$x in K$$, then $${x_n}$$ is a convergent sequence.

My attempt:

Notice that we can define for each $$n$$, $$f_n(x)=d(x_n,x)$$, this is an isometry and thus these are equicontinuous functions.

Also, each converging sequence gives us the fact that these functions are pointwise bounded.

Thus we can us Arzela-Ascoli Therom to show that in fact the sequence of functions $${f_n}$$ has a unifromly convergent subsequence.

Now if I can show that their limit is indeed of the form $$f(x)=d(a,x)$$ for some $$ain K$$, then I can show that $$a$$ is the limit of the sequence, but I don’t know how to do it or even if it is true.

If $$(x_n)_{ninBbb N}$$ diverges, it has subsequences $$(x_{n_k})_{kinBbb N}$$ and $$(x_{m_k})_{kinBbb N}$$ with distinct limits, $$l_1$$ and $$l_2$$ respectively. The fact that it has a convergent subsequence whose limit is some $$l_1$$ follows from the compacity of $$K$$. Since the sequence diverges, there is some $$varepsilon>0$$ such $$d(x_k,l_1)geqslantvarepsilon$$ for infinitely many $$k$$'s. And from those $$x_k$$'s you can extract another convergent subsequence, again by the compacity of $$K$$.

But then the sequence $$bigl(d(x_n,l_1)bigr)_{ninBbb N}$$ diverges, since it has a subsequence whose limit is $$0$$ and another subsequence whose limit is $$d(l_1,l_2)$$.

Correct answer by José Carlos Santos on December 19, 2020

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