# Does the lindelöf space always have countable free sequence?

Does the lindelöf space always have countable free sequence?

A $\kappa$-long free sequence in a space $X$ is a transfinite sequence $S=\{x_\alpha: \alpha < \kappa\}$ of elements of $X$ such that for every $\alpha <\kappa$ the closures in $X$ of the sets $\{x_\beta: \beta < \alpha \}$ and $\{ x_\beta: \alpha \le \beta< \kappa\}$ are disjoint.

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A point is Lindelof. –  Qiaochu Yuan Feb 27 '13 at 6:55
l think he means at most –  Paul Feb 27 '13 at 7:00

(This answer is under the assumption that you are asking whether all free sequences in Lindelöf spaces have countable length; otherwise any length $1$ sequence would be a free sequence of finite — and thus countable — length.)

There are Lindelöf (even compact) spaces with free sequences of uncountable length.

It is an old result of Arhangel'skiĭ that for a compact (Hausdorff) space $X$ the supremum of the lengths of all free sequences in $X$ is equal to the tightness of that space.

(Recall that for a space $X$ and $x \in X$, the tightness of $X$ at $x$, $t ( X,x )$, is defined to be the least cardinal $\kappa$ such that for all $A \subseteq X$ if $x \in \overline{A}$ then there is a $A_0 \subseteq A$ of cardinality $\leq \kappa$ such that $x \in \overline{A_0}$; the tightness of $X$ is then defined to be $t ( X ) = \sup_{x \in X} t(X,x)$.)

It follows that if $\kappa > \omega_1$ is a regular cardinal, then the compact space $\kappa + 1$ has free sequences which are uncountably long (since $\kappa$ would not be a limit point of any $A \subseteq \kappa$ of cardinality $\leq \aleph_1$, but clearly $\kappa \in \overline \kappa$).

Addendum: Of course, if you want specific examples, it is perhaps easier. Suppose that $\kappa$ is any uncountable cardinal, and consider the sequence $\langle \beta + 1 : \beta < \kappa \rangle$ in $\kappa + 1$. Given any $\alpha < \kappa$ it is easy to check that $$\overline{ \{ \beta + 1 : \beta < \alpha \} } = [ 1 , \alpha ]\\ \overline{ \{ \beta+1 : \alpha \leq \beta < \kappa \} } = [ \alpha+1,\kappa).$$ Thus we have a free sequence in the compact space $\kappa + 1$ of length $\kappa$.

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