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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?

Keshav Srinivasan
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  • 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: https://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. – M A Pelto Oct 06 '18 at 04:20
  • @MattAPelto But are there any Cauchy sequences that all topologically equivalent metrics must have in common, apart from convergent sequences? – Keshav Srinivasan Oct 06 '18 at 05:21
  • 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 06 '18 at 10:31
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    @MattAPelto The same is true for uniformly equivalent metrics (which is a weaker requirement than Lipschitz equivalent). – Paul Frost Oct 06 '18 at 11:27
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    @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 06 '18 at 11:51

2 Answers2

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

Alex Kruckman
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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 $d'$.

Eric Wofsey
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