0

First, take some definition: Given a topological space \left ( X,T \right ) and net, defined on X: \left \langle x_{\alpha } \right \rangle_{\alpha \epsilon A}. We say that x is cluster point of \left \langle x_{\alpha } \right \rangle net, if for every U\supset x open set and for every \beta \epsilon A, there exist \gamma \succeq \beta , such that x_{\gamma }\epsilon U.

We know that sequence is a special kind of net. Also, every subsequence is subnet, but all subnet is not a subsequence.

It is known the theorem: Given a arbitrary topological space and net on it. than, x is the cluster point of the net if and only if, there exist subnet of the net, which converges to x.

This theorem is true if the net is sequence (as net). This means, that is there exist subsequence, which is converges to x, than there exist subnet, that converges to x (because subsequence is a subnet) and than x is the cluster point of the sequence (as net). But, conversely we cant say, that if x is the cluster point then there exist subsequence, that is converges (by theorem, where exist subnet but we dont know that this subnet is a subsequence). But it is easy to prove, that if topological space is first countable, than there exist a convergence subsequence also.

I want to find topological space (of cource, it would be not first countable) and sequcence on it, which has a cluster point (x), but there is not subsequence which is converges to x.

VDGG
  • 95
  • I would start with looking at topological spaces that are not first countable. Do you have several examples of those? – uniquesolution Jul 29 '20 at 22:17
  • Simple example is co-final topology (in uncountable set) or co-countable topology, but unfortunetely, they is not work in that sence. – VDGG Jul 29 '20 at 22:23
  • 1
    In the cofinite topology any subsequence is convergent, and the cocountable topology has no non-trivial convergent sequences at all, and moreover these spaces are ugly (non-Hausdorff). – Henno Brandsma Jul 29 '20 at 22:29

1 Answers1

1

In this post I give an argument on why $X=\{0,1\}^I$ is such a space, where $I=\{0,1\}^{\Bbb N}$. This is a huge power of a two point discrete space, so compact Hausdorff and has a sequence without a convergent subsequence even though every net (so every sequence too) has a cluster point, i.e. a convergent subnet. It's a clean(ish) diagonal argument.

$\beta \Bbb N$ (the Cech-Stone compactification of the integers) is another such compact Hausdorff space, where the sequence $x_n =n$ has uncountably many cluster points, but no convergent subsequence at all. It requires a bit more advanced theory though, but if you happen to know the space already..

Henno Brandsma
  • 242,131
  • Thank a lot!!! Easy examples, such as co-finite of co-countable spaces (on uncountable set), does not works. Am I wright? – VDGG Jul 29 '20 at 22:26
  • @VDGG No, we need a bit of work for such spaces. But they do occur naturally. – Henno Brandsma Jul 29 '20 at 22:27
  • 1
    @VDGG The cocountable topology on an uncountable set does have the property that any sequence with all terms distinct has no convergent subsequence. So it's a minimal sort of example (non-compact and non-Hausdorff, while these are both..) – Henno Brandsma Jul 29 '20 at 22:31