Exercise 13, § 8

Let $X$ be a countable Hausdorff space. Suppose that, for each finite sequence $s$ all of whose terms are 0 or 1, there is a non-empty subset $B(s)$ of $X$ which satisfies the following conditions: (i) If $s$ is a finite sequence of $n$ terms (equal to 0 or 1) and $s', s''$ are the sequences of $n + 1$ terms (equal to 0 or 1) whose first $n$ terms are the same as those of $s$, then $B(s') \cup B(s'') = B(s)$ and $B(s') \cap B(s'') = \varnothing$; and if $s_0$ is the empty sequence, $B(s_0) = X$. (ii) For each $x \in X$, if $s_n$ is the (unique) sequence of $n$ terms equal to 0 or 1 such that $x \in B(s_n)$, then the filter base formed by the $B(s_n)$ converges to $x$ in $X$. Show that, under these conditions, $E$ is homeomorphic to the rational line $\mathbf{Q}$. [Show first that $X$ is homeomorphic to a countable subset of Cantor’s triadic set $K$ which is dense in $K$ and contains no end-point of an interval contiguous to $K$. In order to do this, remark that the hypotheses associate with each $x \in X$ an infinite sequence $(u_n(x))$ whose terms are equal to 0 or 1; using the fact that $X$ is countable, show that the bijection $s \to B(s)$ can be modified in such a way that for no $x \in E$ do the $u_n(x)$ form a stationary sequence; finally use Exercise 10 of § 2]. Deduce that every countable subspace of $\mathbf{R}$ which has no isolated points is homeomorphic to $\mathbf{Q}$.