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In many posts on MSE, it is discussed that Cauchy sequences can't be defined in General topological spaces and in a typical topology book it is discussed what converging sequences are, but, what I don't understand is, why, on an abstract level, does convergence generalize even without a metric while cauchy-ness doesn't?

My issue with the posts (eg) is that they show that Cauchyness is not a topological property using specific sequences. But what I wish to know is the abstract idea behind it.

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    You should read about "uniformities" or "uniform structures". This concept is between metrics and topologies and is precisely what one needs to define the Cauchy property for sequences (and nets). – Moishe Kohan Aug 30 '22 at 18:58
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    I think this question is somewhat incorrectly focused. Since Cauchyness in the very definition uses a notion that goes beyond topology (namely distance), we should expect it to not make sense in general topological spaces. The situation that requires explanation is the opposite: when a notion, which on the face of it uses (say) metric ideas, does happen to make sense in a natural way in the looser setting of topological (or even metrizable topological) spaces. – Noah Schweber Aug 30 '22 at 19:09
  • @MoisheKohan: I would not say it is precisely what one needs to define the Cauchy property. The notion of a Cauchy space is what does it, although it goes beyond topological spaces. Maybe all Cauchy topological spaces are uniform, but I don't know that. Do you? (This is a serious question.) – tomasz Aug 30 '22 at 19:57
  • And I think I patched that @NoahSchweber – tryst with freedom Aug 30 '22 at 20:38
  • @tomasz: In your comment, what is a "Cauchy topological space"? As it stands, your comment does not make sense without knowing that. The point of MoisheKohan's comment is that uniform structures were invented for the very purpose of abstracting those additional structural properties, above and beyond a topology, that are needed in order to make sense of Cauchy sequences. – Lee Mosher Sep 01 '22 at 15:54
  • @LeeMosher: I mean a topological space with a compatible Cauchy structure (which should mean that convergent nets are Cauchy, I guess, and maybe some niceness assumptions about the Cauchy structure, like being pretopological, I didn't think too much about it). – tomasz Sep 01 '22 at 21:35
  • No, I am not defining Cauchy. – tomasz Sep 01 '22 at 21:41
  • Thanks for that link @tomasz. – Lee Mosher Sep 02 '22 at 16:06

2 Answers2

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The first abstract idea to come to terms with is this:

The following statement is false: For every topological space $X$, for any two metrics $d,d'$ that generate the topology on $X$, and for every sequence $(x_n)$ in $X$, the sequence $(x_n)$ is a Cauchy sequence with respect to $d$ if and only if it is a Cauchy sequence with respect to $d'$.

In your post you wrote

"... they show that Cauchyness is not a topological property using specific sequences. But..."

... But ... an important thing to realize is that those specific sequences amount to a proof that the above statement is false. All one needs for that proof is to give one counterexample, i.e. one example of $X,d,d',(x_n)$ such that $d,d'$ generate the topology on $X$, and $(x_n)$ is a Cauchy sequence with respect to $d$, and $(x_n)$ is not a Cauchy sequence with respect to $d'$.


With that out of the way, your question regarding the abstract idea behind "Cauchyness" is still very interesting. The comments give two distinct answers to this question: that abstract idea can be uniform structures; or it can be Cauchy structures. One thing I'm unsure of is whether these two notions of "Cauchyness" are equivalent, but I believe they are, in fact I think that this equivalence is covered in this part of the first link above (if I'm wrong about this, I hope someone corrects me).


Okay, so let me try to answer your question by explaining why a uniform structure captures the abstract idea behind Cauchyness. Consider a sequence $(x_n)$ in a topological space $X$.

Assuming that the topology on $X$ is generated by the metric $d$, let recall the standard definition: to say that $(x_n)$ is a Cauchy sequence means that for any $\epsilon > 0$ there exists an integer $N \ge 1$ such that for any integers $m,n \ge 1$ we have $|x_n-x_m| < \epsilon$.

But now, let me reword this definition in a vague and intuitive fashion, without referring to the metric nor to any number $\epsilon$: Given some uniform measurement of closeness in $X$ there exists an integer $N \ge 1$ such that all of the terms of the sequence starting from $x_N$, namely all of the terms in the set $\{x_N,x_{N+1},x_{N+2},...\}$, are simultaneously close to each other with respect to that given uniform measurement of closeness.

What goes wrong in an ordinary topological space $X$ is that there is no such thing as a "uniform measurement of closeness". At best, an open neighborhood of a point $x \in X$ gives a local notion of closeness around $x$, in the sense that being an element of that neighborhood is a notion of "being close to" $x$. But "closeness to $x$" is not (and should not) be transitive! That local notion gives you no way to test whether all the points in an entire infinite set like $x_N,x_{N+1},x_{N+2},...$ are simultaneously close to each other.

The idea of a uniform structure is that a "uniform measurement of closeness" can be expressed by choosing a subset $U \subset X \times X$. Having made a choice of $U$, one can then declare that the points of a set such as $\{x_N,x_{N+1},x_{N+2},...\}$ are all uniformly close to each other if and only if all ordered pairs drawn from that subset are elements of $U$. A uniform structure is then defined to be a set $\Phi$, each of whose elements $U \in \Phi$ is a subset $U \subset X \times X$, such that $\Phi$ satisfies a bunch of axioms.

Here's an exercise: show that if $d$ is a metric on $X$, and if for each $\epsilon > 0$ we define $U_\epsilon = \{(x,y) \in X \times X \mid d(x,y) < \epsilon\}$, then $\Phi = \{U_\epsilon \mid \epsilon > 0\}$ is a uniform structure on $X$.

One can then define a sequence $(x_n)$ in $X$ to be Cauchy if for every $U \in \Phi$ there exists an integer $N \ge 1$ such that every ordered pair drawn from the set of terms $\{x_N,x_{N+1},x_{N+2},\ldots\}$ is an element of the set $U$.

Lee Mosher
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Here's my attempt to explain.

Consider this definition of convergence that does not require a notion of metric:

A sequence $\{x_n\}_{n\in\mathbb N}$ converges to $x$ if and only if for each open neighborhood $U$ of $x$ there is an $m_U \in \mathbb{N}$ such that $x_n \in U$ whenever $n \geq m_U$.

Now, try to imagine an analogous definition for Cauchy sequences. In the convergence definition, we don't need to shift $U$ in anyway. In the Cauchy case, we would have to say something like:

A sequence $\{x_n\}_{n\in\mathbb N}$ is Cauchy if and only if for each open neighborhood U of the origin, there is an $m_U \in \mathbb{N}$ such that $x_j,x_k \in (U + x_m)$ whenever $j,k \geq m_U$.

Hence, the difference is the Cauchy definition has this implicit "shifting" of open sets, and such a mechanism requires an "origin" and a metric.

jacktrnr
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    This, btw, gives rise to a generalization of Cauchy sequences to topological groups: they have their neutral element as the "origin", and shifting is done using the group operation. But I'd like to mention that your definition of Cauchyness should go more like: for every neighborhood of the origin $U$, $x_j-x_j$ is in that neighborhood for sufficiently large $j,k$. You used some unspecified $x_m$ (also sufficiently large?) for your shift, but on a philosophical level, the shift should be by the limit point of the sequence, which may not exist unless the space is complete. – Vercassivelaunos Aug 31 '22 at 02:41
  • How is the "$+$" operator defined in a general topological space? – Lee Mosher Sep 01 '22 at 15:50
  • @LeeMosher Sorry, this is precisely the point I wanted to make -- defining the "$+$" operator requires a metric. – jacktrnr Sep 08 '22 at 14:49
  • This is still not clear. I don't see how to define the expression $U + x_m$ in a metric space. In the very special case of $\mathbb R^n$, sure, it's clear how $+$ is defined. But this question is about general metric spaces. – Lee Mosher Sep 08 '22 at 16:33