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.

  • $\begingroup$ 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. $\endgroup$
    – Thorgott
    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.


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