5
$\begingroup$

Let $X$ be a topological (Hausdorff) space and let $(X_\alpha)_\alpha$ be a directed family of subsets. We say that $(X_\alpha)_\alpha$ generates the topology of $X$ if a subset $U \subseteq X$ is open iff $U\cap X_\alpha$ is open in $X_\alpha$ with respect to the induced topology. Another way of saying this is that $X$ is the direct limit of the topological spaces $(X_\alpha)_\alpha$ where each $X_\alpha$ holds the subspace topology.

It is well-known that the topology of a metrizable space is generated by the family of all compact subsets (one says $X$ is a "$k$-space") since the topology is determined by convergent sequences.

With exactly the same argument, we obtain the result that the topology of a metrizable space is generated by all countable subsets.

My question is now:

For wich non-metrizable Hausdorff spaces it is true that the topology is generated by countable subsets?

Trivially, this also holds for countable spaces but I would expect that there are many more examples. Also, a counter-example would be interesting.

In particular I am interested in examples of the form

$\mathbb R^I := \prod_{i\in I} \mathbb R$ or $(\mathbb Z/2\mathbb Z)^I := \prod_{i\in I} (\mathbb Z/2\mathbb Z) $

for an uncountable index set $I$, but of course all other examples or counter-examples are welcome as well.

Thanks in advance, Tom

$\endgroup$
  • 1
    $\begingroup$ There is some information in en.wikipedia.org/wiki/Countably_generated_space . $\endgroup$ – Emil Jeřábek Nov 4 '14 at 12:35
  • $\begingroup$ thank you, I did not know that these spaces are called like that. $\endgroup$ – Tom Nov 4 '14 at 12:41
  • $\begingroup$ $\mathbb{R}^I$ is countably generated if and only if $I$ is countable. Same thing for $2^I$. $\endgroup$ – Ramiro de la Vega Nov 4 '14 at 13:09
  • $\begingroup$ Thank you very much. This answers my question completely. Do you have a reference for this? $\endgroup$ – Tom Nov 4 '14 at 15:50
  • $\begingroup$ @RamirodelaVega : I recommend that you convert your comment into an answer once you back it up with a reference or an argument. That way Tom can accept it and the question will be marked "answered." $\endgroup$ – Timothy Chow Nov 4 '14 at 16:12
7
$\begingroup$

A space $X$ is said to have countable tightness if whenever $A \subseteq X$ and $p\in \bar{A}$, there is a countable $B \subseteq A$ such that $p \in \bar{B}$. It is not hard to see that a space has countable tightness if and only if its topology is generated by countable sets (in the sense described in the question), but I find it easier to think in these terms.

There are easy examples of non-metrizable countably tight spaces. For instance, the one point compactification of an uncountable discrete space.

Obviously any first-countable space has countable tightness, so both $\mathbb{R}^I$ and $2^I$ have countable tightness if $I$ is countable.

If $I$ is uncountable then let $A$ be the set of functions which take value $1$ at countably-many coordinates and value $0$ at the rest. Then the function with constant value $1$ (call it $p$) is in the closure of $A$ but not in the closure of any countable subset of $A$. Here we are working in either $\mathbb{R}^I$ or $2^I$; so none of these has countable tightness.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.