Q: Is every free sequence discrete? 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.
My question is this:
Is such $\kappa$-long free sequence always discrete whenever for any $\kappa$?

A: Well, almost Yes.
For each $\alpha$, the set $\overline{\{x_\beta\mid \alpha\le\beta<\kappa\}}$ contains none of the $x_\beta$ with $\beta<\alpha$ and if $\alpha+1<\kappa$ then $\overline{\{x_\beta\mid \beta<\alpha+1\}}$ contains none of the $x_\beta$ with $\alpha+1\le\beta<\kappa$, thus their union contains all $x_\beta$ except $x_\alpha$, hence $\{x_\alpha\}$ is relatively open.
But this fails if $\alpha+1=\kappa$.

Hence discreteness follows only if $\kappa$ is a limit ordinal (or for the trivial case $\kappa=1$).

Example: Let $\kappa=2$, $X=\{x_0,x_1\}$ with topology $\{\emptyset, \{x_0\},X\}$.
If we additionally assume $X$ is a $T_1$ space (points are closed), we get a little more:
Assume $\kappa=\alpha+1=\lambda+2$. Then $\overline{\{x_\beta\mid \beta<\lambda\}}$ is a closed set disjoint to $\{x_\lambda,x_\alpha\}$. Since the singleton $\{x_\lambda\}$ is closed, so is $\overline{\{x_\beta\mid \beta<\lambda\}}\cup\{x_\lambda\}$, hence $\{x_\alpha\}$ is relatively open.

Hence for $T_1$ spaces discreteness follows unless $\kappa$ is one more than a limit ordinal.

Example: Let $\kappa=\omega+1$, $X=\mathbb R$, $x_n=\frac 1n$, $x_\omega=0$.

A: Sure.  Given $\alpha$ the complement of the closure of $\{x_{\beta}|\beta<\alpha\}$ is an open set that contains $x_{\alpha}$ and none of the $x_{\beta}$ for $\beta<\alpha$.  Similarly, the complement of the closure of $\{x_{\beta}|\beta\geq\alpha+1\}$ is an open set that contains $x_{\alpha}$ and none of the $x_{\beta}$ for $\beta>\alpha$.  The intersection of these two open sets is an open set that contains $x_{\alpha}$ and no other element of the sequence.
