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What would be your example of an open proper subset $S \subsetneq \mathbb R$ such that $\mathbb{Q} \subsetneq S$?

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$$\exists\text{"set"}\in\text{a question}\not\implies\text{[set-theory] fits.}$$ –  Asaf Karagila Oct 15 '12 at 21:01

2 Answers 2

up vote 12 down vote accepted

$( - \infty , \sqrt{2} ) \cup ( \sqrt{2} , + \infty )$.

Hint for (b): Note that if $( a_n )_{n=1}^\infty$ is an increasing sequence converging to $1$, then for each $x < 1$ there is an $n$ such that $x < a_n$. If you transform this idea into an open cover of $[0,1)$, you will probably find that no finite subcover of this cover can cover $[0,1)$.

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One way of seeing this as an answer is to note that the complement of S has to be a closed set which doesn't contain any rational numbers. Any one will do, so choose the simplest. –  Mark Bennet Oct 15 '12 at 21:07
@ArthurFischer What do you think about the b) part ? –  DreamLighter Oct 15 '12 at 21:27

a) $A_k:=\bigcup_{j=1}^{+\infty}(r_j-2^{-j-k},r_j+2^{-j-k})$ where $\{r_j\}$ is an enumeration of rationals. This show that such that set can be taken as small (in measure) as we want.

b) Take $O_n:=(-1/n,1-1/n)$; then $\{O_n,n\in\Bbb N^*\}$ is an open cover of $[0,1)$ which doesn't have any finite subcover.

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Your open cover is faulty. –  GEdgar Oct 15 '12 at 21:28
@GEdgar You are right. –  Davide Giraudo Oct 15 '12 at 21:31
The new one is OK. –  GEdgar Oct 15 '12 at 21:33

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