What sequences are Cauchy in all metrics for a given topology?

Question

Different metrics for the same topology can have different sets of Cauchy sequences. But I'm interested in what sequences are Cauchy in every metric for a given topology. For a completely metrizable topology, the answer is obvious: the convergent sequences. (Where convergence is a property of the topology independent of the metric.)

But my question is, if a topology is metrizable but not completely metrizable (like $\mathbb{Q}$ with the standard topology), is it possible for non-convergent sequences to be Cauchy in every metric for a given topology?

Answer 1

The sequences which are Cauchy for every compatible metric on $X$ are exactly the (topologically) convergent sequences. Every convergent sequence is easily seen to be Cauchy for every compatible metric, so we're asking about the converse.

So let $(x_n)$ be a sequence in a metrizable space $X$ which is not convergent. Pick a metric $\delta$ on $X$. If $(x_n)$ is not $\delta$-Cauchy, we're done.

Otherwise, if $(x_n)$ is $\delta$-Cauchy, let $\overline{X}$ be the completion of $X$ with respect to $\delta$. Now $(x_n)$ converges to a unique limit point $x\in \overline{X}$, and $\overline{X}$ is completely metrizable. Let $Y = \overline{X}\setminus \{x\}$. We have $X\subseteq Y\subseteq \overline{X}$, and $Y$ is an open subset of $\overline{X}$. It is a theorem that a subspace of a completely metrizable space is completely metrizable if and only if it is $G_\delta$. In particular, there is a compatible complete metric $\delta'$ on $Y$. But $(x_n)$ is not convergent in $Y$, so it is not $\delta'$-Cauchy. The restriction of $\delta'$ to $X$ is a compatible metric in which $(x_n)$ is not Cauchy.

Being a logician, the first reference I can point to for the theorem about $G_\delta$ subspaces is Classical Descriptive Set Theory by Kechris, Theorem I(3.11). But in the case of just removing a single point $y$, it's not hard to write down an explicit $\delta'$ that works: $$\delta'(a,b) = \delta(a,b)+\left|\frac{1}{\delta(a,y)} - \frac{1}{\delta(b,y)}\right|.$$

Answer 2

Here is an alternate proof that a non-convergent sequence is always non-Cauchy in some metric. Let $(X,d)$ be a metric space and $(x_n)$ be a Cauchy but non-convergent sequence in $X$. Note that the set $A=\{x_n\}$ is closed and discrete in $X$, since any accumulation point of $A$ would be a limit of $(x_n)$ (since it is Cauchy) but $(x_n)$ does not converge. Now let $f:A\to\mathbb{N}$ be a bijection ($A$ must be infinite or else $(x_n)$ would converge). By the Tietze extension theorem, $f$ extends to a continuous function $g:X\to\mathbb{R}$, which has the property that $g(x_n)\to\infty$.

We can now define a new metric $d'$ by $d'(x,y)=d(x,y)+|g(x)-g(y)|$. Since $g$ is continuous, $d'$ induces the same topology as $d$. Since $g(x_n)\to\infty$, $(x_n)$ is not Cauchy with respect to $d'$.