Exercise 13, § 8

Let $E$ be a separated countable topological space. Suppose that for every finite sequence $s$ whose terms are equal to 0 or 1, there exists a nonempty subset $B(s)$ of $E$, satisfying the following conditions: 1° If $s$ is a finite sequence of $n$ terms (equal to 0 or 1), $s', s''$ the sequences of $n + 1$ terms (equal to 0 or 1) whose first n terms are equal to those of s, one has B(s') ∪ B(s'') = B(s), B(s') ∩ B(s'') = ∅ ; moreover, if s_0 is the empty sequence, B(s_0) = E.

2° For every x ∈ E, if s_n denotes the (unique) sequence of n terms equal to 0 or 1 such that x ∈ B(s_n), the filter basis formed by the B(s_n) converges to x in E.

Now prove that, under these conditions, E is homeomorphic to the rational line Q. (First establish that E is homeomorphic to a countable subset of the Cantor ternary set K, dense with respect to K and containing no endpoint of an interval contiguous to K. For this, one will note that the hypotheses associate with each x ∈ E an infinite sequence (u_n(x)) whose terms are equal to 0 or 1; using the fact that E is countable, one will show that one can modify the bijection s ↦ B(s) so that for no x ∈ E do the u_n(x) form a stationary sequence. Finally use exerc. 10 of IV, p. 48).

Deduce that every countable subspace of R without isolated point is homeomorphic to Q.