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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?

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  • $\begingroup$ Yes it is. For example in the discrete topological space every constant sequence is a Cauchy one. $\endgroup$
    – Tolaso
    Commented Aug 24, 2015 at 8:22
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    $\begingroup$ 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. $\endgroup$
    – drhab
    Commented Aug 24, 2015 at 8:32
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    $\begingroup$ @Tolaso That makes no sense. It is only an example, and the question is dealing about general topological spaces. $\endgroup$
    – drhab
    Commented Aug 24, 2015 at 8:35
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    $\begingroup$ 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. $\endgroup$ Commented Aug 24, 2015 at 8:43
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    $\begingroup$ @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.) $\endgroup$ Commented Aug 24, 2015 at 8:46

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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.

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    $\begingroup$ Also, the property of being a complete metric space is not invariant under homeomorphism. $\endgroup$
    – Zhen Lin
    Commented Aug 24, 2015 at 8:43
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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.

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    $\begingroup$ 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. $\endgroup$ Commented Aug 24, 2015 at 9:48
  • $\begingroup$ 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. $\endgroup$
    – Siminore
    Commented Aug 24, 2015 at 9:55
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    $\begingroup$ 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). $\endgroup$ Commented Aug 24, 2015 at 10:10
  • $\begingroup$ 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. $\endgroup$ Commented Aug 24, 2015 at 13:15

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