Why closed implies sequentially closed but not the converse?

Question

I'm sure this question has a quick answer but I cannot find it.

According to Wikipedia in a topological space $X$ any closed set $F\subseteq X$ is sequentially closed, but the converse anly holds when $X$ is a sequential space.

But I for the, the following proves that actually sequentially closed implies closed:

Suppose that $F$ is sequentially closed: for any $x\in X$ and any sequence $(x_n)$ on $F$ with limit $x$ then $x\in F$. But if because the definition of a limit point of a sequence , for each neighbourhood of $x$ there exists $n_0\in\mathbb N$ such that $x_n\in V$ whenever $n\geq n_0$. And because the sequence is made of eements of $F$, every neighbourhood of $x$ meets $F$. From this i conclude that every sequentially closed set is closed. Where am I laying?? Which step is false?

Because obviously my reasoning is false: for examplee the sequential topology is finer than the usual topology (I can see that open implies sequentially open). Hence the closures (and thus the closed sets) must be smaller. Also the example given by @Arthur.

Answer 1

The reason that sequentially closed doesn't imply closed is that a sequence has too few points.

The best counterexample I can think of concerns the first uncountable ordinal, usually denoted by $\omega_1$. Under the order topology, $[0,\omega_1)\subseteq [0,\omega_1]$ is sequentially closed, but not closed.

Any convergent sequence of countable ordinals (i.e. any sequence of points in $[0,\omega_1)$) has a countable limit, so $[0,\omega_1)$ is sequentially closed. But the complement of $[0,\omega_1)$ in $[0, \omega_1]$, which is just $\{\omega_1\}$, is not open, so $[0, \omega_1)$ is not closed.

You can fix this discrepancy by using nets rather than sequences. A net is a generalisation of a sequence where you use an arbitrary set with a nice enough order as the index set, rather than limiting yourself to just $\Bbb N$.

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Your flaw is thinking that because $x$ is a limit point of $F$, there is a sequence of points $x_n\in F$ converging to $F$. That's not necessarily the case (as evidenced by my example above).

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Mistake in the new edit: Yes, if $F$ is sequentially closed, it contains all its sequential limit points. So of course, if $x$ is the limit of a sequence in $F$, it is contained in $F$. However, that doesn't show that $F$ is closed, because there may be limit points which aren't sequential limit points. You haven't checked those, and therefore you cannot conclude that $F$ contains them, which again means you cannot conclude that $F$ is closed.

Answer 2

If $x$ is a limit point of set $F$ then every open set containing $x$ will meet $F$, but from that it cannot be concluded that a sequence $(x_n)_n$ in $F$ converging to $x$ exists.

If we are dealing e.g. with cocountable topology then convergence of $(x_n)_n$ to $x$ implies the existence of some $m\in\mathbb N$ that satisfies $n>m\implies x_n=x$ and consequently in that situation every set is sequentially closed. If however the space is uncountable then every uncountable subset that does not equal the whole set is not closed.

Answer 3

Closed implies sequentially closed: if $C$ is closed and $x_n \to x$ in $X$ with all $x_n$ from $C$, then any neighbourhood of $x$ contains points of the sequence, so intersects $C$, so $x \in \overline{C} = C$, and $x$ is sequentially closed.

But if $C$ is sequentially closed, and $x \in \overline{C}$ we only know that every neighbourhood of $x$ intersects $C$, but there could be uncountably many such neighbourhoods so choosing these intersection points does not construct a sequence from $C$ converging to $x$ (which would be needed to show $x \in C$ by sequential closedness). This idea can be made to work in a first countable space (like a metric space, where neighbourhoods of the form "$\frac1n$-balls around $x$" form a countable local base), but not in general as examples show (see @Arthur's post).