Is it possible to define Cauchy sequences in a topological space?

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?

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

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.

• Also, the property of being a complete metric space is not invariant under homeomorphism. Commented Aug 24, 2015 at 8:43

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.

• 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. Commented Aug 24, 2015 at 9:48
• 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. Commented Aug 24, 2015 at 9:55
• 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). Commented Aug 24, 2015 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. Commented Aug 24, 2015 at 13:15