Equivalence of cauchy sequences/completeness in topological spaces.

Question

In general a topological space may not have an equivalent of a metric and its not possible to define completeness of a sequence that way. An alternative that I was thinking about was to say a sequence $x_n$, we say that it is cauchy in the space if there exists a decreasing sequence of open sets $A_n$ such that $x_n\in A_n$ and $\cap A_n $ is $\emptyset$.

This will allow us to construct a topological extension in which there is a point p which is a limit point of $x_n$. The intersection being empty is simply necessary to ensure that the new space is Hausdorff is the old one is.

I would like to know if any of this has been done before by someone or maybe its just not a very fruitful approach. In particular, it might turn out that all sequences are cauchy or atleast a very wide class of them is.

Answer 1

The problem is that this definition doesn't respect metric structure, as Plop points out. But here is a concrete example. Consider $\Bbb N$ with the discrete metric. We have that a sequence is Cauchy (in the metric) if and only if it is eventually constant.

Now let $x_n=n$ and $A_n=\{k\mid k\geq n\}$. Then $A_n$'s are decreasing open sets, with empty intersection, and of course $x_n\in A_n$ as wanted. Of course in the one-point compactification of $\Bbb N$ this is a Cauchy sequence, and it has a limit, but the point of Cauchy sequences is that they use more structure than just the topology to determine when a sequence "should" converge.

On a similar note one can observe the many differences between "complete metric space" and "completely metrizable space" (in the former we are given a metric, in the latter know such metric exists). For example $(0,1)$ is not a complete metric space, but it is completely metrizable as it is homeomorphic to $\Bbb R$. Similarly the irrational numbers are not complete, but they are completely metrizable.

The reason I bring this up is that you can note how with the change of metric, different sequences suddenly become Cauchy, or stop being Cauchy. But the topology stayed the same. This shows up that defining Cauchy-ness of a sequence (or a net) cannot come from the topology alone.

Answer 2

If $E$ is a metric space, and $(x_n)_{n \in \mathbb{N}}$ is a sequence without accumulation points. Then there is a $\varepsilon_0$ such that for all $n>0$, $x_n \not \in B(x_0,\varepsilon_0)$. There is now a $\varepsilon_1$ such that for all $n > 1$, $x_n \not \in B(x_1,\varepsilon_1)$ and $B(x_0,\varepsilon_0) \cap B(x_1,\varepsilon_1) = \emptyset$. Going on and on, you can contruct a family of $\varepsilon_n$, such that $A_n = \bigcup_{m\geq n} B(x_m,\varepsilon_m)$ satisfies the properties you asked.

There is something that can generalize Cauchy sequences, it is called the uniform spaces : http://en.wikipedia.org/wiki/Uniform_space