Proof: Every closed set of real numbers is an $F_\sigma$ set

Question

I've been thinking about the proof. Although it seems very obvious that by the definition of $F_\sigma$ sets, a set is an $F_\sigma$ set if it is a countable union of closed sets, so intuitively, a single closed set is also a collection of '1' closed sets.... But this seems to be a little trivial, it might be that this is actually correct but i am really confused.

Another doubt that came to my mind is whether:' closed set of real numbers' means a collection of singletons?, if yes then would it be different from an 'open set of real numbers'? what is a 'closed set composed of in ANY topological space'? Really new to the subject, trying to get the feel for it..

Answer 1

It is indeed as trivial as you think. A union of one closed set is a countable union of closed sets.

More interestingly: every open set (in $\mathbb{R}$ or any metric space) is also the countable union of closed sets.

Also an open set is a $G_\delta$ (a countable (1) intersection of open sets) and dually a closed set (in a metric space) also is a $G_\delta$ (follows from the open set is an $F_\sigma$ fact by taking complements and applying de Morgan).

Answer 2

A closed set is, by definition, the complement of an open set. A closed set is indeed an $F_\sigma$ set, as it is a countable union of closed sets, exactly as you said!

as for the part about singletons, well singletons are usually not open (for instance in $\mathbb{R}$).

You have the following rules:

• any union of open set is open
• any finite intersection of open sets is open

Hence, you get for closed sets that :

• any intersection of closed sets is closed
• any finite union of closed sets is closed

Answer 3

Let $X$ be a closed subset of reals.

Because $X$ is closed,

$$X=\{x\in\mathbb{R}:(\forall \varepsilon>0)[ X\cap(x-\varepsilon,x+\varepsilon)\ne\varnothing]\}=:\mathrm{closure}(X).$$

That is to say, the set $X$ contains its limit points. So, for example, the interval $(0,1)$ is not closed because it is missing $0$ and $1$---limit points.

Observe $X=X\cup\varnothing\cup\varnothing\cup\cdots$ where the union is countable. Therefore $X$ is an $F_\sigma$-set.

Assume $X=[0,1]$ the closed interval from $0$ to $1$. Then it is true that $X=\bigcup_{x\in X}\{x\}$ however the union here is not countable. Therefore this is not an argument for why $X$ is indeed an $F_\sigma$-set. See the above paragraph for the true argument.