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I am now starting to prepare for a discrete mathematics class. On a test, I came across the following question:

Which of the following sets are countable? $$\mathbb{Z},\mathbb{R}, \mathbb{R-Q}, \{31,2,2019\} $$

The only countable sets are: $\mathbb{Z}$ (easily proved) and $\{31,2,2019\}$ as it is a finite set. Using Cantor's method we prove that there is not a bijective function, such that $\mathbb{R}$ is countable. So there is only $\mathbb{R-Q}$, which is the set of irrational numbers. Can anyone suggest a proper way for me to prove that this is a uncountable set?

average_discrete_math_enjoyer
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5 Answers5

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First prove that the union of two countable sets is countable.

Let $\mathbb{R}\setminus\mathbb{Q}=X$. If $X$ is countable then $\mathbb{R}=X\cup \mathbb{Q}$ is countable, which it isn't.

JMP
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Hint: I believe you know $\mathbb{Q}$ is countable. Suppose that $\mathbb{R}\setminus\mathbb{Q}$ is countable as well. What can we conclude from the assumption that both these sets are countable?

Mark
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$\Bbb{R}$ is uncountable and $\Bbb{Q}$ is countable. If $\Bbb{R}-\Bbb{Q}$ we’re countable the union of two countable sets being countable $\Bbb{R}$ would be countable.

marwalix
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We know \begin{equation} \mathbb{R}=(\mathbb{R}\setminus \mathbb{Q})\cup \mathbb{Q} \end{equation} and $(\mathbb{R}\setminus \mathbb{Q})\cap \mathbb{Q}=\emptyset$. Assume that $(\mathbb{R}\setminus \mathbb{Q})$ is countable. Then taking Lebesgue measure on both sides of the above equality we obtain $$\infty=\mu(\mathbb{R})=\mu((\mathbb{R}\setminus \mathbb{Q})\cup \mathbb{Q})=\mu(\mathbb{R}\setminus \mathbb{Q})+\mu( \mathbb{Q})=0+0=0$$ a contradiction. Hence $\mathbb{R}\setminus \mathbb{Q}$ is not countable.

Note: Here we are using the fact that the Lebesgue measure of any countable set is $0$.

Dastan
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Well, if $\Bbb R\setminus\Bbb Q$ would be countable, then $\Bbb R = (\Bbb R\setminus\Bbb Q) \cup \Bbb Q$ would be the disjoint union of two countable sets and so would also be countable, but it is not.

Wuestenfux
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