Set that is closed under limits but not closed topologically?

Question

I know that in topological spaces that obey the first countability axiom, a set is closed iff any convergent sequence of elements in it that converge converge to within it.

I've been told this isn't true in general: the right-implication yes, but not the left one. I'm looking for an example that shows that it is, in fact, not true in general.

Answer 1

If $X$ is an uncountable set in the co-countable topology, then it obeys the property

If $x_n \to x$ for some sequence in $X$, then there is some $N$ such that for all $n \ge N$, $x_n =x$.

This implies that for all subsets $A$ of $X$ we have that if a sequence from $A$ converges, it converges to an element from $A$. But only $X$ and the at most countable subsets are closed.

Another example: let $X = \mathbb{R}^I$ where $I$ is uncountable, in the product topology. Let $A = \{x \in X: \{i: x_i \neq 0\} \text{ at most countable}\}$, the set of all sequences with so-called countable support.

Then $A$ is dense in $X$, as is easy to check ,as it intersects all non-empty basic open sets, so it's not closed, but it is closed under sequential limits. This is a nice exercise to check.

The classic example, which requires knowing a little about ordinal numbers, is $X = \omega_1 + 1$ in the order topology and $A = \omega_1$ which is not closed but closed under sequence limits, as a countable set of countable ordinals has a countable ordinal as its supremum.

Answer 2

Let S be all the ordinals from 0 to $\omega_1,$ including $\omega_1.$
Though every converging sequence into A = S - {$\omega_1$}
converges to point in A, A is not closed.