MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.
Sign up to join this community
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.
The choice of sequences matters.
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.
