# Is there a compact space with no countably generated dense subspace?

This is a reformulation of this MO question which recieved little or no attention due to the fact that the OP gave no motivation whatsoever. I found the question quite interesting and decided to give it another try (if this is not OK please let me know and I´ll delete this post).

It is known that if $G$ is an abelian compact topological group then it contains a dense subgroup $H$ which is countably tight (in fact Frechet-Urysohn). However the following is open (at least it was a few years ago):

If $G$ is a compact group, must $G$ contain a dense subspace of countable tightness?

This is problem 4.1.1 in "Topological Groups and Related Structures" by A.Arhangelskii and M.Thachenko. Problem 4.1.7 (also open) in the same book is:

Is it true that every homogeneous compact space contains a dense subspace of countable tightness?

My guess is that there should be known examples of (non-homogeneous) compact spaces such that any dense subspace has uncountable tightness, but I could not find any. So I have two questions:

1) Is there such a compact space?

2) For a cardinal $\kappa > 2^{\aleph_0}$ what is a dense subspace of $[0,1]^\kappa$ that has countable tightness?

Perhaps the answer to 2) is that there is none, and $[0,1]^\kappa$ is indeed a counterexample for the second quoted question, but I wouldn´t expect that. Note that if $\kappa \leq 2^{\aleph_0}$ then $[0,1]^\kappa$ is separable and any countable dense subspace would do the trick.

Edit: As Santi suggests in a comment to his answer of 2), the space $X=\beta\mathbb{N} \setminus \mathbb{N}$ is a good candidate for 1), but I still don´t know for sure. Some facts about $X$ that might be relevant: a) $X$ is compact, b) any dense subset of $X$ has size at least $2^{\aleph_0}$, c) the tightness of $X$ is $2^{\aleph_0}$, d) there are $R$-points in $X$, i.e. there exist an open $U \subseteq X$ and a point $x \in \overline{U}$ such that $x \notin \overline{A}$ for any $A \subseteq U$ with $|A|<2^{\aleph_0}$.

• Note: $X$ is countably generated (= has countable tightness) if and only if the closure of any subset $A$ of $X$ equals the union of closures of all countable subsets of $A$. (from Wikipedia) – Gerald Edgar Feb 26 '13 at 17:43

An $\eta_1$-set is a linearly ordered set $(Q, \leq)$ with the property that if A and B are countable subsets of $Q$ satisfying $a < b$ for all $a \in A, b \in B$, then there is an $x \in Q$ satisfying $a < x < b$ for all $a \in A, b \in B$. (This definition and the notation are from Gillman and Jerison.) Let $(X, \leq)$ be the Dedekind compactification of an $\eta_1$-set, that is, $X$ is constructed from the $\eta_1$-set $Q$ in analogy to the construction of the reals from the rationals using Dedekind cuts, and then a first and last element is added. Then $X$ is a connected compact space. I can't recall a reference, but $X$ is the disjoint union of three dense sets: the set of P-points (which includes the endpoints), the set of points that are limits of strictly increasing sequences, and the set of points that are limits of strictly decreasing sequences. Furthermore, in $X$ a point is a limit of a non-trivial sequence if and only if it is an accumulation point of a countable set. If $D$ is a dense subset containing a P-point, it obviously does not have countable tightness. If $D$ is a dense subset containing a limit $x$ of a strictly increasing sequence, then $x$ is not a limit of a strictly decreasing sequence, so the set $S$ of elements of $D$ larger than $x$ contains $x$ in its closure, but no countable subset of $S$ has $x$ in its closure. Similarly, a dense subset containing a limit of a strictly decreasing set does not have countable tightness.
• I guess one can modify this example to get a compact space for which the tightness of any dense subspace is at least $\kappa$ (for any $\kappa$ given in advance). – Ramiro de la Vega Mar 21 '17 at 17:08
Ramiro, regarding 2), the set of all points in $I^\kappa$ with at most countably many non-zero coordinates is even Frechet-Urysohn, for every $\kappa$. This is due to Noble (see also exercise 3.10.D in Engelking).
Note: Frechet-Urysohn means that given a non-closed set $A$ and a point $x \in \overline{A} \setminus A$, there is a sequence inside $A$ converging to $x$.