# Is every rational sequence topology homeomorphic?

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?

• At least a couple correct answers were given here on MathOverflow. May 26, 2023 at 18:18

I don't know if for different choices of sequences the spaces can still be homeomorphic, but from the definition, we can see that the topology definitely does depend on the choice of the sequences. By axiom of choice, for every irrational $$x$$, choose $$u_n(x)$$ a sequence approaching $$x$$ from below and $$v_n(x)$$ from above.

For the first topology, the interval $$(-\infty, x]$$ is a neighborhood of $$x$$, but it isn't a neighborhood for the second topology.

I would suspect that there might be some homeomorphism, very likely constructed by axiom of choice, but from this example we can see that the neighborhoods definitely depend on the choice of the sequences.

• Yeah, these are good observations, but it's quite possible that $f(x)=-x$ could be the homeomorphism in your example. May 13, 2023 at 18:19

Saúl on MathOverflow gave this answer:

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.