Uncountable Set with Co-countable topology is not a Sequential Space

Question

[Citation: S. Morris "Topology without Tears" problem 6.2.11.iii]

Here's the problem:

"A topological space $(X,\tau)$ is said to be a sequential space if every sequentially closed set is closed. Prove that a topological space is a sequential space if and only if every sequentially open set is open. Deduce from this that if $X$ is an uncountable set and $\tau$ is the countable-closed topology on $X$ of exercise 1.3 #6, then $(X,\tau)$ is not a sequential space."

I get the first proof, but I'm not sure about the "deduce" part. It seems to suggest finding a seq. open set that is not open, but I don't see how to do it that way. I think I'm missing something and/or making it more difficult than it needs to be.

Here's my attempt:

If we can find a seq. closed set that is not closed, then $X$ cannot be a seq. space.

Fix an element $a\in X$, and consider the proper subset $S=X\setminus \{a\}$

Since $X$ is uncountable, we know $S$ is also uncountable.

Since $\tau$ is the co-countable topology, we know that $S$-uncountable is not closed.

But $S$ is sequentially closed by the following argument:

For any element $b\in S$, consider the sequence $S_b = \{b, b, b, ...\}$. Clearly, $S_b \rightarrow b\in S$.

Now, show that $a\not\in S$ is not the limit of any sequence in $S$:

Note that any sequence $S_n$ of elements of $S$ is a countable set, and let $M = \{x: x = x_n$ for some $x_n\in S_n\}$ Therefore, the set $X\setminus M$ is open.

But note also that since $a\not\in S$, we have that $a\in X\setminus M$ for all possible sequences $S_n$.

That means that for any sequence $S_n = \{x_1, x_2, x_3, ...\}$ taken from $S$, there exists an open set $X\setminus M$ containing $a$ but not containing any of the sequence elements.

(*) By the definition of convergence of sequence, we say $S_n\rightarrow a$ if for each open set $U$ containing $a$ there exists a positive integer $N$ such that $\forall n\geq N$, $x_n\in U$.

But $X\setminus M$ is an open set containing $a$ that contains none of the elements in the sequence $S_n$.

Therefore, there can be no positive integer $N$ such that $\forall n \geq N$, $x_n\in X\setminus M$.

Since this is the case for every sequence in $X\setminus\{a\}$, there is no sequence $S_n$ taken from $S$ such that $S_n \rightarrow a$.

Therefore, $S$ is sequentially closed.

But we saw that $S$ is not closed.

So, we have found a seq. closed set that is not closed. Therefore, $(X,\tau)$ is not a sequential space.

Answer 1

Lemma: if $X$ has the property that all countable sets are closed and $x_n \to x$ in $X$ then there is some $N$ such that for all $n \ge N$ we have $x_n =x$. (i.e. all convergent sequences are eventually equal to their limit).

Proof: Suppose $x_n \to x$. Define $M=\{x_n: x_n \neq x\}$. This is a countable set that does not contain $x$, so $X\setminus M$ is an open neighbourhood of $x$ so by convergence there is some $N$ such that for all $n \ge N$, $x \in X\setminus M$. The latter can only occur if $x_n =x$ (by definition) so we are done.

This means that in the co-countable topology all subsets are sequentially closed: let $A \subseteq X$ and suppose that $a_n \in A$ for all $n$ and $a_n \to x$. By the lemma we have that for some $N$, $a_n = x$ for all $n \ge N$ so in particular $x \in A$. So $A$ is sequentially closed.

Now let $A$ be any uncountable proper subset of $X$ (so your $X$ minus a singleton will do). This is not closed (it's not $X$ or countable) but it is sequentially closed by the previous. Done.