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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 (*)?

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  • $\begingroup$ See en.wikipedia.org/wiki/Uniform_space $\endgroup$ – Daniel Schepler Oct 5 '18 at 23:34
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    $\begingroup$ 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. $\endgroup$ – Andreas Blass Oct 5 '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.

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    $\begingroup$ 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". $\endgroup$ – btilly Oct 5 '18 at 23:45
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    $\begingroup$ $x_n = (-1)^n n$, $U_n = (-\infty, -n) \cup (n, \infty)$? $\endgroup$ – Daniel Schepler Oct 6 '18 at 0:00
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    $\begingroup$ $x_n = (-1)^n (1 - \frac{1}{n})$, $U_n = (-1, -1 + \frac{1}{n}) \cup (1 - \frac{1}{n}, 1)$. $\endgroup$ – Daniel Schepler Oct 6 '18 at 0:09
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    $\begingroup$ @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. $\endgroup$ – Alex Kruckman Oct 6 '18 at 14:58
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    $\begingroup$ To characterize completeness, you must refer to some other structure beyond the topology alone, like a uniform structure or a metric. $\endgroup$ – Alex Kruckman Oct 6 '18 at 15:00
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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.

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  • $\begingroup$ 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 $\bigcup_{n}U_n = \{0\}$. $\endgroup$ – Alex Kruckman Oct 6 '18 at 17:26
  • $\begingroup$ @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. $\endgroup$ – Paul Frost Oct 7 '18 at 13:11

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