# Proving that $\mathbb{Q}$ is neither open or closed in $\mathbb{R}$

I need to figure out if $\mathbb{Q}$ is either open or closed in $\mathbb{R}$ and prove my assertions. I don't think it's either. I would like a review/critique of my proofs to make sure that there's not an easier way to do it.

To see that it's not open, take some $\frac{m}{n}$ with both $m$ and $n$ perfect squares so that $\left(\frac{m}{n}\right)^{\frac{1}{2}}$ is rational. We have that $x \mapsto x^{\frac{1}{2}}$ is continuous. Hence for any $\epsilon$-neighborhood of $\left(\frac{m}{n}\right)^{\frac{1}{2}}$, there exists a $\delta$-neighborhood of $\frac{m}{n}$ that maps into it. but

$$\begin{eqnarray*} \frac{m}{n} - \frac{cm}{cn + 1} < \delta &\Leftrightarrow& \frac{m}{n(cn + 1)} < \delta n(cn + 1) \\ &\Leftrightarrow& \frac{m}{\delta n^{2}} < \left(nc^{2} + 1\right)^{2} \\ &\Leftrightarrow& \frac{m}{\delta n^{4}} < c^2 + \frac{2}{n}c + \frac{1}{n^{2}} \\ &\Leftrightarrow& \frac{m}{\delta n^{4}} < \left(c + \frac{1}{n}\right)^{2} \\ &\Leftrightarrow& \left(\frac{m}{\delta n^{4}}\right)^{\frac{1}{2}} - \frac{1}{n} < c \end{eqnarray*}$$

So we can make $\frac{cm}{cm + 1}$ as close to $\frac{m}{n}$ as we like by taking $c$ sufficiently large. If $cm$ is a perfect square, then $c$ must be a perfect square because $m$ is so we can just choose a $c$ that's not a perfect square and hence $\left(\frac{cm}{cn +1}\right)^{\frac{1}{2}}$ is irrational. Because we can place an irrational arbitrarily close to at least one positive rational, $\mathbb{Q}$ is not open.

To show that $\mathbb{Q}$ is not closed, simply pick some irrational $\alpha$ and let $\epsilon_n = \frac{1}{n}$. Let $x_n$ be chosen from between $\alpha - \epsilon_n$ and $\alpha$. Then the sequence $x_n$ converges to $\alpha$ which shows that the complement of $\mathbb{q}$ in $\mathbb{R}$ is not open as we can place a rational in any $\epsilon$-neighborhood of any irrational.

Proving that the rationals aren't open was so much harder than proving that they're not closed because my book hadn't proved that the irrationals are dense in $\mathbb{R}$ and so I couldn't use that. Is there some other way that I could do this?

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assume it 's closed, find a rational sequence whose limit is an irrational number. – Yuan Apr 2 '11 at 5:33
assume it's open, then its complement is closed, then find an irrational sequence whose limit is a rational number – Yuan Apr 2 '11 at 5:34
@Yuan: Those initial assumptions seem superfluous, turning a direct argument (or contraposition) into proof by contradiction. – Jonas Meyer Apr 2 '11 at 5:37
I don't understand why you need to go through the rigamarole of using the square root function. Just note $\sqrt{2}/n$ can be made arbitrarily small, so $q\pm\sqrt{2}/n$ is an irrational that is arbitrarily close to a rational $q$... – Arturo Magidin Apr 2 '11 at 5:42
@Arturo: Could you please explain to me what you did in your editing the question? I couldn't find the offending piece of code and then I saw someone else editing and figured it was you, so I gave up. – t.b. Apr 2 '11 at 5:45

## 3 Answers

You are certainly right that all you need to show that $\mathbb{Q}$ is not open is to show that given a rational $q$, you can find irrationals arbitrarily closed to $q$. A very simple way of doing this is to take your favorite irrational, say $\sqrt{2}$, and then consider $\sqrt{2}/n$ with $n$ an integer. These numbers are never rational, and get arbitrarily small; so if $q$ is a given rational, then $q+\frac{\sqrt{2}}{n}$ will be as close as you want to $q$ by taking $n$ large enough.

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There are many proofs of this.

NOT CLOSED:

• The rationals are dense because the reals are the Cauchy completion of the rationals, so $\mathbb{Q}$ is not closed.
• The rationals are dense because the reals are the Dedekind completion of the rationals, so ditto.
• The decimal expansion of $\pi$ gives a sequence of rationals whose limit is not rational, so again it's not closed.
• Any closed dense set contains a copy of the Cantor space, which has size continuum.

NOT OPEN:

• Singletons are closed, so if the rationals were open, each set of the form $\mathbb{Q} \setminus \{ q \}$ would be open dense, so their intersection will be dense by Baire category, but their intersection is empty.
• Open sets have positive measure, but the rationals have measure 0.
• The irrationals are dense: If $q$ is some rational, then the irrational sequence $\pi ^{1/n}q$ tends to $q$. So does $q - \frac{\pi}{n}$. Dense sets must meet every open set but the rationals don't meet the irrationals, so the rationals aren't open.
• Also since the irrationals are dense, they're not closed or else they would be all of $\mathbb{R}$. Hence their complement, the rationals, aren't open.
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You can prove that the irrational numbers are dense, and then you will be able to use it. If $\alpha$ is any irrational number, then the nonzero rational multiples of $\alpha$ form a dense set of irrationals.

E.g., to show that there is a rational multiple of $\sqrt{2}$ in the interval $(a,b)$, you could first find a nonzero rational $r$ in the interval $\left(\frac{a}{\sqrt 2},\frac{b}{\sqrt 2}\right)$, then multiply by $\sqrt 2$. (You want $r\neq 0$ so that $r\sqrt 2$ is irrational.)

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