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I know that we can define Cauchy sequences in topological vector spaces. How about in general topological spaces? Is it possible to define a Cauchy sequence in general topological spaces?

Leeroy Jenkins
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  • Yes it is. For example in the discrete topological space every constant sequence is a Cauchy one. – Tolaso Aug 24 '15 at 08:22
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    I think that the space must at least be metrizable. This to make it possible to define a Cauchy sequence. And that is probably not all. Wich metric must be elected? Thinking like this I arrive at metric spaces. – drhab Aug 24 '15 at 08:32
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    @Tolaso That makes no sense. It is only an example, and the question is dealing about general topological spaces. – drhab Aug 24 '15 at 08:35
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    What you need is a "uniform" concept of nearness in the space. A topology is not strong enough for that, a uniform structure is what you need to define Cauchy sequences (Cauchy filters/Cauchy nets). Note that a vector space topology defines a uniform structure. The fact that the "nearness" of $x$ and $y$ can be considered as the "nearness" of $x-y$ and $0$ gives the uniformity. – Daniel Fischer Aug 24 '15 at 08:43
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    @drhab We need less than a metric. A uniform structure is enough. (Aside: every uniform structure can be defined by a family of semimetrics, so we could also say we need a semi-metrisable space.) – Daniel Fischer Aug 24 '15 at 08:46
  • @DanielFischer Thank you for your educational comments. It confirms that my decision to comment (not to answer) and to write "I think" was a correct one. I am not familiar yet with the stuff you mention. – drhab Aug 24 '15 at 08:52
  • The idea of defining a cauchy sequences(which is not well defined) in a topological space is similar to defining a ball in a topological space. –  Aug 24 '15 at 16:17

2 Answers2

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No. Consider $X=(0,1)$ and $Y=(1,\infty)$ equipped with the usual metric. These are homeomorphic as topological spaces, since the map $h:X\to Y$, defined by $$h(x)=\frac1x$$ is a homeomorphism. But $h$ maps the Cauchy sequence $a_n=\frac1n$ to $h(a_n)=n$, which is not a Cauchy sequence. So being a Cauchy sequence is not invariant under homeomorphisms, but depends on the choice of a metric.

Dejan Govc
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In general topological spaces Cauchy sequences are not defined. Let us think of a possible definition. In metric spaces, we all know the definition, and we could try to mimic it. However, what is the topological counterpart of "$d(p_n,p_m)<\varepsilon$"? We could try

Definition. A sequence $\{p_n\}_n$ is a Cauchy sequence if, for every open set $U$, there exists $N>1$ such that $p_n$ and $p_m$ belong to $U$ for all $n$, $m>N$.

But this definition does not mean that $p_n$ and $p_m$ are as "close" as we wish when $n$ and $m$ become large: already in $\mathbb{R}$, pick $U=(0,1) \cup (100,1000)$. What is required in the definition of Cauchy sequences is some kind of "uniform neighborhood". And indeed Cauchy sequences are defined in topological vector spaces and in topological uniform spaces.

tryst with freedom
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Siminore
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    I don't get your definition at all. It seems to be saying a sequence is Cauchy if it converges to every point in the space (and even then you would need to exclude $U=\emptyset$), which usually is impossible. Certainly you meant to say something different. – Marc van Leeuwen Aug 24 '15 at 09:48
  • No, I meant exactly this: in a topological space you can't replace a uniform neighborhood (or a neighborhood of zero in topological vector spaces) by a "generic neighborhood" and obtain a reasonable definition. – Siminore Aug 24 '15 at 09:55
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    The notion "generic neighborhood" cannot be "open set", since the former notion is relative to a given point (of which it is a neighborhood), while the latter is not. Certainly your $U$ needs to be qualified in some way (require it so contain some given point). But anyway, what you wrote does not look like the definition of a Cauchy sequence at all (even if you assume a topological vector space). – Marc van Leeuwen Aug 24 '15 at 10:10
  • Perhaps a more useful-seeming statement of that definition would be: “A sequence ${p_n}_n$ is a Cauchy sequence if for every $L\geq 1$, for every open set $U\ni p_L$, there exists $N>L$ such that $p_n$ and $p_m$ belong to $U$ for all $n, m>N$”. Of course that's in fact just as useless. – leftaroundabout Aug 24 '15 at 13:15