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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.

One Answer

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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