Dtyjlui

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 $yin X$ , then since the function $ f: X to mathbb R $ defined as $f(x)=d(x,y), forall xin X$ 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.

Let's define $a_n = lim_{mto infty} d(x_n, x_m)$. 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 $a_{k_0} = lim_{mto infty} d(x_{k_0}, x_m) in (frac{z}{2},frac{3z}{2})$, 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 $x_n$.

Whether the metric $d$ is complete or not, if $Y$ is a non-empty subset of $X$ then $d(x,Y)=inf {d(x,y):yin Y}$ 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 $inf {d(a,b): ain Aland bin B}>0.$ So let $f(x)=d(x, {x_n:nin A})$. Then $f(x_n)=0$ when $nin A$ but $inf {f(x_n):nin B}>0$ so ${f(x_n)}$ is not convergent.