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I was given this assertion in class and told that the proof is really short. I understand that for an open interval that is bounded such as $(a,b)$ we can define a closed set $[a-\frac{1}{n},b+\frac{1}{n}]$ with $n \in \mathbb{N}$. However, I'm not sure how to extend this to an unbounded interval.

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possible duplicate of Are all compact sets in $ \Bbb R^n$, $G_\delta$ sets? – William Sep 15 '12 at 5:14
@William I wouldn't say this is a duplicate, since not all open sets are compliments of compact sets. – Alex Becker Sep 15 '12 at 5:43
@AlexBecker The answers prove that closed sets are $G_\delta$. – William Sep 15 '12 at 5:44
Alternatively, make use of the distance function to the complement $F = U^c$: let $d_F(x) = \inf_{f \in F} d(x,f)$, so $d_F(x)$ measures how far $x$ is from $F$ and it is a continuous function. Then the fact that $F$ is closed yields that $d_F(x) = 0$ if and only if $x \in F$ implies that $$U = \bigcup_{n=1}^\infty \left\{x : d_F(x) \geq \frac{1}{n}\right\}$$ which exhibits $U$ as a countable union of closed sets. This doesn't use any special property of the real numbers beyond existence of a distance. See here for some facts on the distance function. – t.b. Sep 15 '12 at 10:12
up vote 3 down vote accepted

For unbounded intervals such as $( a , + \infty )$ you just have to look at the bounded side: for each $n$ let $F_n = [ a + \frac{1}{n} , + \infty )$, which is closed, and show that $( a , + \infty ) = \bigcup_{n=1}^\infty F_n$. (And analogously for $( - \infty , b )$.)

To get that every open set in $\mathbb{R}$ is F$_\sigma$, note the following:

  1. Every open set in $\mathbb{R}$ is the (disjoint) union of countably many open intervals.
  2. Every open interval in $\mathbb{R}$ is F$_\sigma$.
  3. It is not too hard to prove that countable unions of F$_\sigma$ sets are also F$_\sigma$.
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So for $3.$ could use say let $X$ be a $F_\sigma$ set such that $X=\cup Y$ where $Y$ is a countable union of closed sets. Then just take the $\cup X$ to show that it is still a countable union of closed sets? – Fmonkey2001 Mar 23 at 5:39

Sets of the forms $[a,\to)$ and $(\leftarrow,a]$ are closed, and

$$(a,\to)=\bigcup_{n\in\Bbb Z^+}\left[a+\frac1n,\to\right)\;.$$

Don’t forget, though, that it’s not enough just to show that every open interval is an $F_\sigma$; you’ve got to show it for all open sets. You’ll need to use the fact that every non-empty open set in $\Bbb R$ can be written as the union of at most countably many pairwise disjoint intervals. (I’m including the possibility of unbounded open intervals.) If you already know this fact, there’s very little left to do; if not, you should prove it, which does take a bit of work.

In case you do have to prove it, here’s a pretty big hint. Let $U$ be a non-empty open set in $\Bbb R$, and define a relation $\sim$ on $U$ by $x\sim y$ if $x\le y$ and $[x,y]\subseteq U$, or $y\le x$ and $[y,x]\subseteq U$. Show that $\sim$ is an equivalence relation and that its equivalence classes are pairwise disjoint open intervals whose union is $U$. Then use the fact that $\Bbb R$ is separable to show that there are at most countably many equivalence classes.

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Is there any reason you use $[a, \to)$ instead of $[a, \infty)$? I have never seen the former notation before. – Michael Albanese Sep 15 '12 at 5:24
@Michael: Very simply, because I prefer it. It suggests the concept very clearly, and it doesn’t suggest that there’s some nebulous quasi-object $\infty$ hanging off the end of the order. In my experience it’s in quite common use amongst people who work a lot with linear orders, though it’s rarely seen in undergraduate texts. – Brian M. Scott Sep 15 '12 at 5:28
Is it automatic that $\mathbb{R}$ and $\emptyset$ are $F_\sigma$ because they are both closed (and open). – emka Sep 17 '12 at 20:44

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