Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an exercise from a topological book. It is this:

Show that any open subsets of the real line are $F_\sigma$-sets.

Could anybody help to solve it?

share|improve this question
1  
You could do it for metric spaces in general. Let $F$ be a non-empty closed set, then $x \mapsto d(x,F)$ is a continuous function vanishing exactly on $F$. It follows that $F$ is a $G_\delta$, being the intersection of countably many open sets: $F = \bigcap_{n=1}^\infty \{x\,:\,d(x,F) \lt 1/n\}$. Thus $F^c$ is an $F_\sigma$. –  t.b. Aug 7 '12 at 7:57
    
@t.b. Very nice! –  Paul Aug 7 '12 at 8:00
add comment

2 Answers

up vote 2 down vote accepted

First note that open subsets of the real line are countable unions of (pairwise disjoint) open intervals. Next show that every open interval is an F$_\sigma$-set. As countable unions of F$_\sigma$-sets are also F$_\sigma$, this gives the result.

share|improve this answer
    
My idea is this: for an open interval $[a,b]$, there are countably open intervals such that their union is $[a,b]$, and hence we may get countably closed set which are the closues of the open intervals and their union is also $[a,b]$. Am I right? –  Paul Aug 7 '12 at 8:10
    
@John: In you comment, is $[a,b]$ an open interval or a closed interval? –  Arthur Fischer Aug 7 '12 at 19:11
    
Sorry. $[a,b]$ should be $(a,b)$. –  Paul Aug 8 '12 at 12:05
    
@John: Don't worry about "closures of open intervals," just worry about closed intervals: if $(a,b)$ is any open interval, then it is a countable union of (non-degenerate) closed intervals (which happen to be the closures of their interiors, but that is not too important). –  Arthur Fischer Aug 8 '12 at 13:45
add comment

Hint: start with an open interval. Can you show it is the union of countably many closed intervals?

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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