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I am confused by this question. Why can't we just have a disjoint union of open sets of which every set is non empty?

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canseeker
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Hint: Let $R>0$. Show that a collection of disjoint open sets in $[-R,R]^n$ has only countably many nonempty sets.

Severin Schraven
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    Am sorry for my unrefined and silly logic. Am a social science major trying to self taught some proper mathematical logic to myself.

    Can the intuition to the hint be :- if set is non empty then it does have a point of Q^n in its interior. If we have non-countable many such open and disjoint sets then measure goes to infinity. Which means even though the union is a subset of R^n its measure equals to \infty ?

     – canseeker May 04 '16 at 16:08
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    @canseeker The $\Bbb Q^n$ argument is the way to go. There are only countably many rational points and every non-empty open set of a collection of disjoint open sets has such a point "of its own". Hence there can only be countably non-empty such sets. – Hagen von Eitzen May 04 '16 at 16:25
  • many thanks to everyone for the help! – canseeker May 04 '16 at 16:28
  • @HagenvonEitzen great argument. I actually had a measure theoretic argument in mind, but yours is way better. – Severin Schraven May 04 '16 at 16:41
  • @SeverinSchraven Could you please share it? – canseeker May 04 '16 at 16:47
  • In fact, you gave the argument yourself. Namely that an infinite collection of disjoint nonempty open set would have infinite measure, which contradicts the fact that $[-R,R]^n$ has finite measure. Now if you have a collection of disjoint open sets $(U_{\lambda}){\lambda \in \Lambda}$ on the whole of $\mathbb{R}^n$, write $\Lambda_R:={ \lambda \in \Lambda : U{\lambda} \cap [-R,R]^n \neq \emptyset }$. By the above reasoning, we get that $\Lambda_R$ is countable. But the set of all the nonempty sets of your collection is $\bigcup_{R\in\mathbb{N}} \Lambda_R$ and therefore countable. – Severin Schraven May 04 '16 at 17:21