# If $X$ is Hausdorff and $|X|> \mathfrak{c}$, does $X$ always have a uncountable discrete subspace?

Let $X$ be a Hausdorff topological space with $|X|> \mathfrak c$. Does $X$ always have a uncountable discrete subspace? Thanks for your help.

• I have downvoted because this question does not meet the quality standards that many users desire for questions on this site: it does not have any background information on where you encountered the problem, nor any information on what you have tried. – Carl Mummert May 1 '13 at 19:06

A result of Hajnal and Juhasz says the following:

Given any Hausdorff space $X$, $|X| \leq 2^{2^{s(X)}}$; in particular every Hausdorff space with countable spread has size $\leq 2^{2^\omega}$.

However, Todorcevic gave the following consistency result:

Assuming PFA, every Hausdorff space with countable spread has size $\leq 2^\omega$.

For the flip side we get the following:

It is consistent that there is a Hausdorff (even collectionwise normal) space with countable spread of size $2^{\omega_1}$.

(This is obtained by adding $\omega_1$ many Cohen reals to a model of CH which additionally satisfies that there is a family of $2^{\omega_1}$ many uncountable subsets of $\omega_1$ such that the intersection of any two distinct members is countable; note that CH will still hold in the extension.)

See the following references for more details:

• R. Hodel, Cardinal functions I, in the Handbook of Set-Theoretic Topology, pp.1-61. (In particular section 5.)
• I. Juhasz, Cardinal functions II, in the Handbook of Set-Theoretic Topology, pp.63-109. (In particular section 2.)