Can a non-perfect-Polish topology have the same "sequential convergence structure" as a perfect Polish topology?

Question

Given a topological space $X$, by its sequential convergence structure I mean the full information about the convergent sequences in $X$ together with their limits. (I guess to be formal, it is the subset of $X^{\mathbb{N}} \times X$ consisting of all pairs $( ( x_n )_{n=0}^\infty , x )$ such that $x = \lim_{n \to \infty} x_n$.)

It is well-known that a topology is not in general characterized by its sequential convergence structure. For example, if you give an uncountable set $X$ both the discrete and the co-countable topologies, then the convergent sequences are exactly the eventually constant sequences which have the obvious limit.

Other examples can be found in the answers to the following question:

• Is the topology that has the same sequential convergence with a metrizable topology equivalent as that topology?

Recall that a topological space $X$ is called Polish if it is separable and completely metrizable. It is called perfect Polish if, in addition, it has no isolated points.

The discrete topology on $\mathbb{N}$ is a common example of a Polish space, however it is very far from being perfect Polish (since all points are isolated). This brings me to my question:

Question: Are there examples of topologies $\mathcal{O}_p$, $\mathcal{O}_n$ on a set $X$ such that

1. $( X , \mathcal{O}_p )$ is a perfect Polish space, and
2. $( X , \mathcal{O}_n )$ is not a perfect Polish space, and
3. $( X , \mathcal{O}_p )$ and $( X , \mathcal{O}_n )$ have the same "sequential convergence structure"?

Answer 1

Consider the usual topology on $[0,1]$, and the topology $T$ which consists of the empty set together with all cocountable sets that are open in the usual topology. Obviously $T$ is not perfect Polish (it is not even Hausdorff). However, I claim that the convergent sequences for $T$ are the same as for the usual topology. Since $T$ is contained in the usual topology, one direction is trivial.

For the nontrivial direction, suppose $(x_n)$ does not converge to $x$ in the usual topology. Then there is some open interval $U$ around $x$ and some subsequence $(x_{n_k})$ of our sequence which is never in $U$. Passing to a further subsequence, we may assume that $(x_{n_k})$ converges to some point $y\in[0,1]$ in the usual topology. Now let $A=\{y\}\cup\{x_{n_k}:k\in\mathbb{N}\}$ and $V=[0,1]\setminus A$. Then $V\in T$, and infinitely many terms of $(x_n)$ are not in $V$. Since $x\in V$, this means $(x_n)$ does not converge to $x$ with respect to $T$.

(Of course, there's nothing special about $[0,1]$ here: you could do the same thing starting with any nonempty perfect Polish space $X$ in place of $[0,1]$. If $X$ is not compact, then $(x_{n_k})$ may not have a convergent subsequence, but then you can just take $A=\{x_{n_k}\}$.)

Answer 2

The answer to your question is that yes, there are such spaces, but I don't know any easy example.

Consider the space $\ell^1$ with its usual topology and with the weak topology. The former is perfect Polish, while the latter isn't metrizable, however $\ell^1$ has the so called Schur's property, meaning that a sequence converges weakly iff it converges in norm (for a proof see Conway's A course in functional analysis, theorem V.5.2, this is a nontrivial result)