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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 $T_f=T_{f'}$ if $f\neq f'$. As there are $2^{2^{\aleph_0}}$ choices for $f$, we are done.

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.

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 $A_{f(x)}$. We cannot have $T_f=T_{f'}$ if $f\neq f'$ (consider the closures of $A_0,A_1$ in $(\mathbb{R},T_f)$), and there are $2^{2^{\aleph_0}}$ functions $\mathbb{R}\setminus\mathbb{Q}\to\{0,1\}$.

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 $T_f=T_{f'}$ if $f\neq f'$. As there are $2^{2^{\aleph_0}}$ choices for $f$, we are done.

There are lots of classes of 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 these spaces is induced by some bijection $\mathbb{Q}\to\mathbb{Q}$. So if all these spaces where homeomorphic, we could only have continuum many rational sequence topologies in $\mathbb{R}$.

However we can construct more: 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 rational sequence topology $T_f$ in $\mathbb{R}$ such that the sequence of rationals convergent to any irrational $x$ is contained in $A_{f(x)}$. We cannot have $T_f=T_{f'}$ if $f\neq f'$ (consider the closures of $A_0,A_1$ in $(\mathbb{R},T_f)$), and there are $2^{2^{\aleph_0}}$ functions $\mathbb{R}\setminus\mathbb{Q}\to\{0,1\}$.

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 $A_{f(x)}$. We cannot have $T_f=T_{f'}$ if $f\neq f'$ (consider the closures of $A_0,A_1$ in $(\mathbb{R},T_f)$), and there are $2^{2^{\aleph_0}}$ functions $\mathbb{R}\setminus\mathbb{Q}\to\{0,1\}$.