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Definition. A subset of $\mathbb{R}$ is called open if it is a union of open intervals.

Example. Show that $\mathbb{Q}$ is not open.

How can I show to use the definition? Can you give a hint?

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    every open ball has irrational numbers – Alladin Jan 13 '17 at 00:29
  • @Alladin Can you explain that why every open balls has irrational numbers? –  Jan 13 '17 at 00:36
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    For example, $a$ and $b$ real numbers, $ I = (a,b)$. We know that $\sqrt{2}$ is irrational, find $p$ and $q$ integers numbers such that $a <\frac{p}{q}\sqrt{2} < b$, note that $\frac{p}{q}\sqrt{2}$ is an irrational number. – Alladin Jan 13 '17 at 00:49
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    @Alladin Thanks. –  Jan 13 '17 at 00:52

2 Answers2

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Hint: Suppose that $\mathbb Q$ is a union of open intervals; let $I$ be one of them and let $a,b \in I$. Now construct an irrational number $x$ such that $a < x < b$, so that $x \in I$, a contradiction.

Théophile
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  • But $x\not\in \mathbb{Q}$, a contradiction. –  Jan 13 '17 at 00:35
  • @Kahler Yes, that is the contradiction: on the one hand, $x \not \in \mathbb Q$, but on the other hand, $x \in I \subset \mathbb Q$. – Théophile Jan 13 '17 at 00:39
  • $I$ is subset of $\mathbb{R}$, is it? –  Jan 13 '17 at 01:04
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    @Kahler Yes. To write out my hint more explicitly: Suppose that $\mathbb Q$ is a union of open intervals of $\mathbb R$; let $I$ be one of them (so $I \subset \mathbb R$), ... – Théophile Jan 13 '17 at 03:29
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Suppose that $\mathbb{Q}$ is open, so $\mathbb{Q}^c$, the set of irrational numbers, is closed. Consider the sequence $(x_{n})_{n}$ given by $x_{n} = \frac{\sqrt{2}}{n}$. Note that $x_{n} \in \mathbb{Q}^c$ for all $n$, and $x_{n} \to 0$, but $0 \notin \mathbb{Q}^c$, contradiction.

Juniven Acapulco
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Alladin
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