*Crossposted from [Math.SE 4698387](https://math.stackexchange.com/questions/4698387/).*




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In the [rational sequence topology](https://topology.pi-base.org/spaces/S000057/properties), 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?



Is every rational sequence topology homeomorphic?
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.


