Cofinite topology (on an uncountable set $X$) and sequential continuity

Question 10C #4 in Willard's General Topology asks which of the properties from Theorem 10.5 hold for an uncountable set $$X$$ endowed with the cofinite topology.

Theorem 10.5 states: Let $$X,Y$$ be first countable spaces. Then,

1. $$U \subseteq X$$ is open iff whenever $$x_n \to x \in U$$, then $$(x_n)$$ is eventually in $$U$$.
2. $$F \subseteq X$$ is closed iff whenever $$(x_n)$$ is contained in $$F$$ and $$x_n \to x$$, then $$x \in F$$.
3. $$f:X \to Y$$ is continuous iff whenever $$(x_n) \to x \in X$$, then $$f(x_n) \to f(x) \in Y$$.

So, first of all, I believe that Property 2 holds (from a previous question I asked here Counter-example: a topology that is not first countable where elements in the closure are exactly the elements that are limits of sequences?). Some googling also brought me to the fact that $$(X, \tau_{cofinite})$$ is a sequential space (terminology we have not covered in my course, but as far as I can tell, it means that Property 3 holds as well).

Based on my further reading of sequential spaces, then Property 1 should hold as well. Another user asked a related question (here: Is a cofinite topological space a sequential space?) which seems to indicate it does hold. However, like the OP of that question, I remain unconvinced that the answer given makes sense and so now I am confused about which properties hold. I suspect I'm misunderstanding something, but want to explain my reasoning.

I think that Property 1 does not hold, due to the following:

Let $$X$$ be an uncountable set, endowed with the cofinite topology. Let $$\alpha \in X$$ and define $$A := \{\alpha\} \subset X$$. Define the constant sequence $$x_n = \alpha$$, $$\forall n \in \mathbb{N}$$. Then $$x_n \to \alpha \in A$$, $$(x_n) \subseteq A$$, but $$A$$ is closed, not open. So doesn't that mean Property 1 doesn't hold?

Any clarification is appreciated.

• I don't understand your suppose counter-example. You would need to show that for all sequence $(x_n)_n$ converging to $\alpha$, they eventually lie in $A$, not just for one specific such sequence. Oct 19 '20 at 0:13

No, that example does not show that Property 1 fails. Property 1 says that $$U$$ is open iff every sequence converging to a point of $$U$$ is eventually in $$U$$.
Let $$\langle x_n:n\in\Bbb N\rangle$$ be any sequence of distinct points of $$X\setminus\{\alpha\}$$; then every open nbhd of $$\alpha$$ contains all but finitely many points of the sequence, so the sequence converges to $$\alpha$$, but it is never in the set $$A$$. (In fact such a sequence converges to every point of $$X$$.) The existence of this sequence shows that $$A$$ is not open.