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

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..

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).

• Thanks, Could you help me with the second part of the question as well? – Cosmic Jan 26 at 11:57
• @Cosmic a closed set of real numbers is a set that is the complement of an open set. Or equivalently a set that contains all its limit points. In defining a topology one defines the open sets and the closed sets are just their complements. – Henno Brandsma Jan 26 at 12:11

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

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.