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Let $A$ be a subset of real numbers containing all the rational numbers. Which of the following statements is true?

a. $A$ is countable.

b. If $A$ is uncountable then $A=R$.

c. If $A$ is open, then $A=R$.

d. None of the above statement is true.

(a) can't be true since $A$ containing all the rational numbers no way implies that $A$ contains only the rational numbers.

The confusion lies between (b) and (c).

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up vote 3 down vote accepted

Hint What if $A = \mathbb{R}-\{\sqrt{2}\}$?

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a. is false since a subset of the real numbers containing the rationals is all of the reals.

b. Wrong. Take $\mathbb{R} - \{\sqrt{2}\}.$

c. Enumerate the rationals and enclose the $n$th in an interval of length $\epsilon/2^n$. This kills c.

Hence, d.

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notice your example on b also eliminates c – clark Feb 4 '14 at 2:15
but the example kills it more spectacularly, with an open set of measure $\le\epsilon$. Bang. dead. – ncmathsadist Feb 4 '14 at 2:24
to be honest, when reading the problem your solution is what came to my mind. Since this is a clean measure theoretic idea. But when I saw the hint of Habert, I said wait a minute this answers only $b$? and then realised of course not it asnwers $c$ as well. So I commented just in case this happened to someone else. – clark Feb 4 '14 at 2:38

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