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I have heard it said that completeness is a not a property of topological spaces, only a property of metric spaces (or topological groups), because Cauchy sequences require a metric to define them, and different metrics yield different sets of Cauchy sequence, even if the metrics induce the same topology. But why wouldn't the following definition of Cauchy sequence work?

(*) A sequence $(x_n)$ in a topological space $(X,\tau)$ is Cauchy if there exists a nested sequence of open sets $(U_n)$ where each open set is a proper subset of the one before it and the intersection of all of them contains at most one element, such that for any natural number $m$, there exists a natural number $N$ such that for all $n\geq N$, we have $x_n\in U_m$.

My question is, why isn't this definition equivalent to being Cauchy in all metrics on $X$ whose topology is $\tau$? Are there any metrics where being Cauchy is equivalent to (*)?

Keshav Srinivasan
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  • See https://en.wikipedia.org/wiki/Uniform_space – Daniel Schepler Oct 05 '18 at 23:34
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    The map $x\mapsto1/x$ is a homeomorphism from $[1,\infty)$ onto $(0,1]$ (both regarded as metric spaces with the usual metric $|x-y|$ from $\mathbb R$). So these two spaces have exactly the same topological properties. But $[1,\infty)$ is complete and $(0,1]$ isn't. So completeness is not a topological property. – Andreas Blass Oct 05 '18 at 23:47

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According to your proposed definition, the sequence $1,2,3,\dots$ would be Cauchy in $\mathbb{R}$, witnessed by the sequence of open sets $U_n = (n,\infty)$.

Edit: Let me incorporate some information from the comments to make this a more complete answer.

As another example of why your definition is unsatisfactory: In $\mathbb{R}^2$, any sequence $(a_n,0)$ of points on the $x$-axis is Cauchy, witnessed by the sequence of open sets $\mathbb{R}\times (-1/n,1/n)$.

The fact that completeness and Cauchyness are not topological properties can be formalized by the fact that there are generally many different metrics compatible with a given topology, and these different metrics can induce different notions of completeness and Cauchyness. Looked at a different way, homeomorphisms preserve all topological properties (I would take this to be the definition of a topological property), but homeomorphisms between metric spaces do not necessarily preserve completeness (see the examples in btilly's and Andreas Blass's comments).

On the other hand, the notion of completeness actually lives somewhere in between the world of topological properties and metric properties, in the sense that many different metrics can agree about which sequences are Cauchy. It turns out that they agree when they induce the same uniform structure on the space. And indeed, completeness can be defined purely in terms of the induced uniform structure, so it's really a uniform property.

There is one class of spaces in which topological property and uniform properties coincide: a compact space admits a unique uniformity. So you could call completeness a topological property for compact spaces. But in a rather trivial way, since every compact uniform space is automatically complete.

It could be worthwhile to view compactness as the proper topological version of completeness, i.e. the topological property that comes the closest to agreeing with the uniform/metric property of completeness.

Alex Kruckman
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    And note that there is a homeomorphism between $\mathbb{R}$ and $(0, 1)$ so they have the same topology. The topology alone therefore doesn't capture whether a particular sequence is "converging". – btilly Oct 05 '18 at 23:45
  • @btilly Thanks, the comparison of the two makes things especially clear. – Keshav Srinivasan Oct 05 '18 at 23:54
  • @Alex, can you give an example of a sequence in $\mathbb{R}$ whose limit isn't infinity or negative infinity but which is still Cauchy under my definition but not under the usual definition? – Keshav Srinivasan Oct 05 '18 at 23:55
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    $x_n = (-1)^n n$, $U_n = (-\infty, -n) \cup (n, \infty)$? – Daniel Schepler Oct 06 '18 at 00:00
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    $x_n = (-1)^n (1 - \frac{1}{n})$, $U_n = (-1, -1 + \frac{1}{n}) \cup (1 - \frac{1}{n}, 1)$. – Daniel Schepler Oct 06 '18 at 00:09
  • @DanielSchepler Hmm... Just trying things out, but what if you require the open sets in the sequence to be connected? – Keshav Srinivasan Oct 06 '18 at 02:55
  • @KeshavSrinivasan why bring in connectedness? There are lots of spaces with outconnected open sets at all. This would limit the definition severely. – Henno Brandsma Oct 06 '18 at 04:55
  • @HennoBrandsma Well, even getting a metric-independent definition of Cauchy sequences that works for spaces with connected open sets would be an accomplishment. – Keshav Srinivasan Oct 06 '18 at 05:19
  • Cauchyness is a uniform space feature. It never depends only on the topology, except for compact spaces; these have a unique compatible uniformity. – Henno Brandsma Oct 06 '18 at 05:20
  • @HennoBrandsma OK, but I’m still trying to find counterexamples to my proposed definition(s). – Keshav Srinivasan Oct 06 '18 at 12:14
  • @KeshavSrinivasan you have plenty of examples already.. – Henno Brandsma Oct 06 '18 at 12:17
  • @HennoBrandsma Do you know of any example of a sequence of real numbers whose limit is not infinity or negative infinity and which is Cauchy under my definition using connected open sets, but not Cauchy under the usual definition? – Keshav Srinivasan Oct 06 '18 at 12:19
  • @KeshavSrinivasan There is no such example in $\mathbb{R}$. That's because a connected open set is just an interval, and the only way a nested sequence of intervals can have an intersection which is empty or just one point is if their lengths decrease to $0$ (which would force the sequence to be Cauchy) or if they all have endpoint $\infty$ or $-\infty$ (which would force the sequence to grow or decrease without bound). – Alex Kruckman Oct 06 '18 at 14:01
  • But all this is extremely special to $\mathbb{R}$. For example, in $\mathbb{R}^2$, look at the sequence of (connected) open sets $(-\infty,\infty)\times (-1/n,1/n)$. Then absolutely any sequence of points on the $x$-axis $(a_n,0)$ is Cauchy with respect to your definition... – Alex Kruckman Oct 06 '18 at 14:03
  • @AlexKruckman OK thanks, the $\mathbb{R}^2$ example makes things clear. Now I’m wondering, what if in my definition I replaced the nested sequence of open sets with a nested sequence of compact sets? Then would there be any sequences Cauchy under my definition but not under the usual definition? – Keshav Srinivasan Oct 06 '18 at 14:47
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    @KeshavSrinivasan With the new definition, you have a problem in the other direction. The sequence $(1/n)$ in the metric space $(0,1)$ is Cauchy by the metric definition, but it doesn't satisfy your new definition with compact sets! Is it clear to you that this definition could not possibly give a correct characterization of completeness? The definition is phrased just in terms of topological notions (e.g. compact sets), so it's preserved by homeomorphisms, but completeness is not preserved by homeomorphisms in general. – Alex Kruckman Oct 06 '18 at 14:58
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    To characterize completeness, you must refer to some other structure beyond the topology alone, like a uniform structure or a metric. – Alex Kruckman Oct 06 '18 at 15:00
  • @AlexKruckman Yeah, I’m aware of that fact, I just find it counterintuitive, so I’m trying to investigate why various attempts to define it topologically fail. – Keshav Srinivasan Oct 06 '18 at 15:43
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You are interested in metrizable spaces $X$ and ask how the set $\mathfrak{C}(X)$ of sequences $(x_n)$ in $X$ which are Cauchy with respect to all metrics on $X$ which induce the given topology $\tau$ on $X$ can be characterized. I guess you want to find a simple property expressed directly via $\tau$.

As you mentioned in What sequences are Cauchy in all metrics for a given topology? $\mathfrak{C}(X)$ contains all convergent sequences, and if $X$ is completely metrizable, then $\mathfrak{C}(X)$ does not contain any other.

As pointed out by the comments and Alex Kruckman's answer to your present question, your definition (*) does not work to characterize $\mathfrak{C}(X)$.

(*) contains two cases:

(a) $\bigcap_n U_n = \emptyset$.

(b) $\bigcap_n U_n = \{ x \}$.

Case (a) describes non-convergent sequences, but the known examples show that such sequences may be Cauchy with respect to one metric, and non-Cauchy with respect to another metric.

Case (b) covers the sequences which converge to $x$ although there may be also non-convergent sequences satisfying (b) (see Alex Kruckman's comment). Only if the $U_n$ form a neighborhood base of $x$ this is excluded. However, if you take any neigboorhood base $U_n$ of $x$ and any sequence $V_n$ satisfying (a), then $W_n = U_n \cup V_n$ again satisfies (b) which shows you the problem.

Paul Frost
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  • Even when $\bigcap_n U_n = {x}$, you can get sequences which don't converge to $x$. For example, let $U_n = (-1/n,1/n)\cup (n,\infty)$. Then $1,2,3,\dots$ is "Cauchy" with respect to these open sets, even though $\bigcup_{n}U_n = {0}$. – Alex Kruckman Oct 06 '18 at 17:26
  • @AlexKruckman Thank you for drawing my attention to that point. I erroneously took it for granted that the $U_n$ form a neigbhorhood base of $x$. I shall edit my answer. – Paul Frost Oct 07 '18 at 13:11