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

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:

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"?

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 $$A=\{x_{n_k}\}$$.)
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)