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.

In the reals with the usual topology, prove or disprove: Every open set is a $G_\delta$ set.

Having real trouble with this, I can prove the other way but I can't seem to get my head around this direction. Any help would be much appreciated!

share|improve this question
    
A side note, a Gδ set is the intersection of countably many open sets however I am guessing this is common knowledge. –  katherinebarry Feb 20 '13 at 16:45
    
You can prove the other way? It is not true that a $G_\delta$ set is open (e.g.: $\{0\}=\bigcap_{n=1}^\infty (-1/n,1/n)$). –  David Mitra Feb 20 '13 at 16:50
    
Your example above is dealing with an infinite number of sets whereas $G_\delta$ is the intersection of countably many open sets –  katherinebarry Feb 20 '13 at 19:36
    
"countable" means "finite or countably infinite". (In fact, some take "countable" to mean "countably infinite".) –  David Mitra Feb 20 '13 at 19:40
    
well then i do apologise for the flaws in the terminology used, but I am sure that the question should be worded in terms of a finite number of open sets. sorry :) –  katherinebarry Feb 20 '13 at 19:44

2 Answers 2

Hint: $A$ is $G_\delta$ if and only if it is a countable intersection of open sets. If $U$ is open, can you find a countable sequence of open sets whose intersection is $U$?

share|improve this answer
1  
hint to the hint: a sequence need not have distinct elements... –  Henno Brandsma Feb 20 '13 at 17:35
1  
Alternative hint to the hint: $1$ is a countable cardinal number. –  Brian M. Scott Feb 21 '13 at 13:26

Hint: For any set $X$ and any subset $Y\subseteq X$, what is $Y\cap Y$?

share|improve this answer
    
So do you mean that it follows trivially - as in we can write each open set as simply intersections of themselves? –  katherinebarry Feb 20 '13 at 19:38
    
@katherinebarry: Yup, that's right! –  Zev Chonoles Feb 20 '13 at 20:51

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.