# Is every rational sequence topology homeomorphic?

Crossposted from Math.SE 4698387.

In the rational sequence topology, rationals are discrete and irrationals have a local base defined by choosing a Euclidean-converging sequence of rationals and declaring any cofinite subset of this sequence along with the irrational to be open.

Do these choices of sequences matter? Or does there exist a homeomorphism for any pair of sequence assignments?

There are lots of classes of RSTs (rational sequence topologies) in $$\mathbb{R}$$ up to homeomorphism. Note that, as mentioned in the answer by Will Brian, any homeomorphism $$(\mathbb{R},T_1)\to(\mathbb{R},T_2)$$ between RST spaces is induced by some bijection $$\mathbb{Q}\to\mathbb{Q}$$. So any homeomorphism class of RST topologies can have at most $$2^{\aleph_0}$$ elements.
However there are $$2^{2^{\aleph_0}}$$ RSTs: let $$A_0,A_1$$ be two disjoint, dense subsets of $$\mathbb{Q}$$, and let $$f:\mathbb{R}\setminus\mathbb{Q}\to\{0,1\}$$ be an arbitrary function. We can construct a RST $$T_f$$ in $$\mathbb{R}$$ such that the sequence of rationals convergent to any irrational $$x$$ is contained in $$A_{f(x)}$$. In the space $$(\mathbb{R},T_f)$$, the closure of $$A_i$$ is $$A_i\cup f^{-1}(i)$$, showing that $$T_f \not= T_{f'}$$ if $$f\neq f'$$. As there are $$2^{2^{\aleph_0}}$$ choices for $$f$$, we are done.
• In particular, $cl_{T_f} A_0 = A_0 \cup f^{\leftarrow}[\{0\}]$ is distinct for each $f$, showing these topologies are all distinct. May 26 at 15:08
The trick is to notice that if $$X$$ is such a space, then the copy of the rational numbers inside of $$X$$ -- let's denote it $$\mathbb Q_X$$ -- is a countable dense subset of $$X$$. This implies that if $$X$$ and $$Y$$ are any two such spaces, any homeomorphism $$X \rightarrow Y$$ is induced by a bijection $$\mathbb Q_X \rightarrow \mathbb Q_Y$$.
Using this observation, we can build, via a transfinite recursion of length $$\mathfrak{c}$$, two such spaces that are not homeomorphic. Begin with two copies $$\mathbb R_X$$ and $$\mathbb R_Y$$ of $$\mathbb R$$, and an enumeration $$\langle f_\alpha :\, \alpha < \mathfrak{c} \rangle$$ of all bijections $$\mathbb Q_X \rightarrow \mathbb Q_Y$$ (of which there are $$\mathfrak{c}$$-many, because $$\mathbb Q_X$$ and $$\mathbb Q_Y$$ are countable). At stage $$\alpha$$ of the recursion, suppose we've already chosen some sequences for your topologies on $$X$$ and $$Y$$, but fewer than $$\mathfrak{c}$$ sequences have been chosen in total. Then we can choose finitely many sequences of rationals converging to irrationals in $$\mathbb R_X$$ and $$\mathbb R_Y$$ in such a way that the map $$f_\alpha$$ cannot induce a homeomorphism between $$X$$ and $$Y$$. (In detail, either (1) $$f_\alpha$$ does not extend to a homeomorphism $$\mathbb R_X \rightarrow \mathbb R_Y$$, this is witnessed by a point whose sequences are not yet chosen, and we can choose a converging sequence in one space that does not map to a converging sequence in the other. Otherwise, (2) $$f_\alpha$$ does represent a homeomorphism of the reals, or at least it does on all the points we have left to consider, but we can choose a sequence converging in $$X$$ to some irrational $$r$$, and then choose in $$Y$$ a sequence converging to the image of $$r$$ under that homeomorphism that is disjoint from the image of the sequence we chose in $$X$$.) By the end of the recursion, we've chosen $$\mathfrak{c}$$ rational sequences for each space and killed all possible homeomorphisms between them. I didn't set up the recursion in such a way that we've necessarily chosen a sequence for every irrational yet, but the remaining choices (if there are any) can be made arbitrarily.