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

[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.

• You have to be a bit more careful in showing that $X\setminus \{a\}$ is sequentially closed. You have to consider any sequence in it and show it can only converge to a point unequal to $a$. A sequence is also not the same as the set of its values. See my answer for a clean approach. Dec 30, 2018 at 8:12
• I see that I was being sloppy in my notation and not distinguishing between a sequence and the set of its values. But I think I actually have considered any seq in X\{a} and shown it cannot converge to a ("That means that for any sequence..."). I'll add a clarification at (*). Dec 31, 2018 at 5:30

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.