On a special type of sequence in complete metric space

Question

Let $\{x_n\}$ be a sequence in a complete metric space $X$ such that for every uniformly continuous function $f:X \to \mathbb R$, the sequence $\{f(x_n)\}$ is convergent in $\mathbb R$. Then is it true that $\{x_n\}$ is convergent ?

My try: If $\{x_n\}$ has a convergent subsequence, then $\{x_n\}$ converges. Indeed, if $\{x_{k_n}\}$ is a convergent subsequence with limit $y\in X$ , then since the function $ f: X \to \mathbb R $ defined as $f(x)=d(x,y), \forall x\in X$ is uniformly continuous, so $\lim f(x_n)=\lim d(x_n,y)$ exists, but since $\lim d(x_{k_n},y)=0$, so $\lim d(x_n,y)=0$, thus $\{x_n\}$ converges to $y$.

Unfortunately, I am unable to decide whether $\{x_n\} $ has a convergent (or for that matter, just Cauchy, since $X$ is complete) subsequence or not.

Answer 1

Let's define $a_n = \lim_{m\to \infty} d(x_n, x_m)$. Since distance is uniformly continuous, each of the $a_n$ exists. Eventually all points are contained within a ball of radius $a_1+\epsilon$ about $x_1$. This implies that the $a_n$ are eventually bounded above by $2(a_1+\epsilon)$, and so by Bolzano-Weierstrass must converge to some nonnegative number $z$. If $z$ is equal to $0$ we can show that the $x_n$ are Cauchy. Otherwise, if $z>0$ then there exists $k_0$ so that $a_k, k \geq k_0$ is within $\frac{z}{2}$ of $z$.

Since $a_{k_0} = \lim_{m\to \infty} d(x_{k_0}, x_m) \in (\frac{z}{2},\frac{3z}{2})$, we can now pick $k_1 > k_0$ so that $d(x_{k_0},x_{k_1}) \in (\frac{z}{2},\frac{3z}{2})$. Continue this process inductively, at each step picking a $k_j$ so that $x_{k_j}$ is at least $\frac{z}{2}$ from all the previously chosen $x_{k_i}$. This gives us a countable subcollection of the $\{x_i\}$ who are mutually at least $\frac{z}{2}$ distance apart.

We can construct a uniformly continuous function $f$ equal to $1$ on $x_{k_i}$ for even $i$ and $0$ on $x_{k_i}$ for odd $i$. This is because all the $x_{k_i}$ are suitably spaced out (as a sketch, we can define $f$ equal to $1$ on the ball of radius $\frac{z}{4}$ about $x_{k_i}$ for even $i$, and have $f$ decrease to zero in a controlled manner by the time we hit the boundary of the radius $\frac{z}{2}$ ball about $x_{k_i}$). This $f$ is equal to $0$ and $1$ infinitely often, contradicting the fact that $z>0$. So $z=0$, and we can pick some convergent subsequence amongst the $x_n$.

Answer 2

Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=\inf \{d(x,y):y\in Y\}$ is uniformly continuous. If $\{x_n\}$ is not a Cauchy sequence then there exists a disjoint pair $A, B$ of infinite subsets of $\Bbb N$ such that $\inf \{d(a,b): a\in A\land b\in B\}>0.$ So let $f(x)=d(x, \{x_n:n\in A\})$. Then $f(x_n)=0$ when $n\in A$ but $\inf \{f(x_n):n\in B\}>0$ so $\{f(x_n)\}$ is not convergent.