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

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?

• If two metrics $d_1$ and $d_2$ on a common set $X$ induce the same topology $(X,\tau)$ then the metrics are topologically equivalent. Unfortunately this tells us nothing but there is a stronger (implies the aforementioned but not vice versa) notion of equivalent metrics: proofwiki.org/wiki/Definition:Lipschitz_Equivalence/Metrics If two metrics are lipschitz equivalent, then a sequence is cauchy with respect to one metric if and only if it is cauchy with respect to the other. I am not too well versed on metrization, so I am unsure when these equivalences might coincide. – Matt A Pelto Oct 6 '18 at 4:20
• @MattAPelto But are there any Cauchy sequences that all topologically equivalent metrics must have in common, apart from convergent sequences? – Keshav Srinivasan Oct 6 '18 at 5:21
• Even in completely metrizable spaces there may be sequences which are Cauchy with respect to one metric, but non-Cauchy with respect to other metrics. This comes from the fact that not all metrics need to be complete. – Paul Frost Oct 6 '18 at 10:31
• @MattAPelto The same is true for uniformly equivalent metrics (which is a weaker requirement than Lipschitz equivalent). – Paul Frost Oct 6 '18 at 11:27
• @PaulFrost Yes, different metrics for a completely metrizable topology may have different sets of Cauchy sequences. But what I said is still true: the set of all sequences which are Cauchy in ALL the metrics for a completely metrizable topology is precisely the set of convergent sequences. I want to know if that’s also true for non-completely metrizable topologies. In other words, I’m interested in the intersection of all the sets of Cauchy sequences for different metrics for the topology. – Keshav Srinivasan Oct 6 '18 at 11:51

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: $$\delta'(a,b) = \delta(a,b)+\left|\frac{1}{\delta(a,y)} - \frac{1}{\delta(b,y)}\right|.$$
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 $$d'$$.