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).

  • $\begingroup$ Thanks, Could you help me with the second part of the question as well? $\endgroup$ – Cosmic Jan 26 at 11:57
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.