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



When is the topology generated by countable subsets?
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. 


