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Let $(S, d)$ be a complete metric space and $(x_n)_{n\in \mathbb{N}}$ a sequence in $S$. If $(f(x_n))_{n\in \mathbb{N}}$ converges for all continuous $f:S\to\mathbb{R},$ does it follow that $(x_n)_{n\in \mathbb{N}}$ converges with respect to $d$?

Intuitively, I think the implication is true, but I can't prove it (can't find a counterexample either).

Any ideas?

rmdmc89
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    @NickE there's no reason that $S$ need be a subset of $\mathbb R$, so the identity function need not be a map from $S$ to $\mathbb R$. – qaphla May 20 '16 at 01:32
  • but $S$ may be totally different from $\mathbb{R}$, so $x\in S$ doesn't imply $x\in\mathbb{R}$ – rmdmc89 May 20 '16 at 01:34

1 Answers1

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This is true. In particular, consider any sequence $x_n$ such that all continuous functions $f:S\rightarrow \mathbb R$ converge. We will start by showing that if $x_n$ has a limit point $y$, then it converges to $x$. In particular, the condition of being a limit point may be expressed as $$\liminf_{n\rightarrow\infty}d(x_n,y)=0$$ however, since $x\mapsto d(x,y)$ is a continuos map the limit of it applied to $x_n$ must exist and therefore coincide with the limit infimum, giving $$\lim_{n\rightarrow\infty}d(x_n,y)=0$$ meaning $x_n$ converges to $y$.

Next, we will show by contradiction that $x_n$ has a limit point. Suppose it didn't. Note two things: First, the image of the sequence $X=\{x_n\}$ is infinite, as no point can appear infinitely often. Second, any subset $S\subseteq X$ is closed, as any point on the boundary of $S$ is either in $S$ or is a limit point of $x_n$. Knowing this, let us take any subset $S$ of $X$ such that both $S$ and $X\setminus S$ are infinite.

At this moment, you could apply the Tietze extension theorem to a continuous function $f$ which is $1$ on $S$ and $0$ on $X\setminus S$ and note that $f(x_n)$ would not converge. If you want to work constructively, then enumerate $S$ as $\{s_1,s_2,\ldots\}$. Define $\varepsilon_n=\min(2^{-n},\inf\{d(s_n,y):y\in X\setminus \{s\}\})$. This is always positive since $X\setminus \{s\}$ is closed and doesn't contain $s$. Now, define the function $$f(x)=\sum_{n=1}^{\infty}\max\left(1-\frac{1}{\varepsilon_n}d(x,y_n),0\right)$$ The reader may check that this is continuous - the trick to showing this is noting that, for any $x$, there is some ball around it in which only finitely many of the summands are non-zero. Moreover, the sum is zero on $X\setminus S$ and one on $S$, so $\lim_{n\rightarrow\infty}f(x_n)$ fails to exist. This is the contradiction we were seeking. Thus we conclude that $x_n$ must have a limit point and therefore must converge.

Milo Brandt
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