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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!

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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

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 $U$?

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hint to the hint: a sequence need not have distinct elements... – Henno Brandsma Feb 20 '13 at 17:35
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 $Y\cap Y$?

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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

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