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

  • 1
    $\begingroup$ There is some information in en.wikipedia.org/wiki/Countably_generated_space . $\endgroup$ Nov 4, 2014 at 12:35
  • $\begingroup$ thank you, I did not know that these spaces are called like that. $\endgroup$
    – Tom
    Nov 4, 2014 at 12:41
  • $\begingroup$ $\mathbb{R}^I$ is countably generated if and only if $I$ is countable. Same thing for $2^I$. $\endgroup$ Nov 4, 2014 at 13:09
  • $\begingroup$ Thank you very much. This answers my question completely. Do you have a reference for this? $\endgroup$
    – Tom
    Nov 4, 2014 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." $\endgroup$ Nov 4, 2014 at 16:12

1 Answer 1


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.


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