How to prove closure of $\mathbb{Q}$ is $\mathbb{R}$?

Can it be proved easily?

  • 1
    $\begingroup$ How are you defining $\mathbb{R}$? Often it is defined as the closure of $\mathbb{Q}$. The answer will give some idea what techniques are allowed. $\endgroup$ – Ross Millikan Jan 26 '11 at 6:43

Presumably, you mean the closure of $\mathbb{Q}$ in $\mathbb{R}$ under the usual topology.

It suffices to show that for every real number $r$ and every $\epsilon\gt 0$, there is at least one rational $q$ which is "$\epsilon$-close" to $r$ (that is, $|r-q|\leq \epsilon$), since this will show that every open ball around $r$ contains a rational. This shows that the complement of $\mathbb{Q}$ has empty interior, so the closure of $\mathbb{Q}$ is all of $\mathbb{R}$.

To prove that, let $r$ be a real number, and let $\epsilon\gt 0$. Then there exists a natural number $n$ such that $\frac{1}{10^n}\lt \epsilon$. Write $r$ in decimal expansion, and let $q$ be the rational that has the same decimal expansion as $r$ up to the $n$th digit after the decimal point, and terminates there. This is a rational number, since its decimal expansion terminates.

Now note that $|r-q|\leq \frac{1}{10^n} \lt \epsilon$, proving that there is a rational that is $\epsilon$-close to $r$.

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    $\begingroup$ How does it follow that if every open ball around $r$ contains a rational, then $\mathbb{Q}$ has empty interior? @ArturoMagidin $\endgroup$ – sequence Feb 16 '16 at 5:46
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    $\begingroup$ @sequence: "then the complement of $\mathbb{Q}$ has empty interior." Because no point has an open ball around it that is completely contained in the complement. $\endgroup$ – Arturo Magidin Nov 17 '17 at 20:38

This question really does not make a whole lot of sense. In topology closures only makes sense in some ambient space. Some examples will probably make this clear.

  • $\mathbb{Q}$'s closure in $\mathbb{Q}$ is $\mathbb{Q}$ itself.
  • $\mathbb{Q}$'s closure in $\mathbb{R}$ is $\mathbb{R}$.
  • $\mathbb{Q}$'s closure in $\mathbb{Q}(\sqrt{2})$ is $\mathbb{Q}(\sqrt{2})$.

And this doesn't even go into embeddings of $\mathbb{Q}$ in e.g. the unit circle. What you probably meant to ask was how to prove that the completion of $\mathbb{Q}$ is $\mathbb{R}$. For that you should take a look at this question.

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    $\begingroup$ Well, perhaps the OP meant to ask how to show that the closure of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$. (But you're right; it's not so clear, and other interpretations are possible.) $\endgroup$ – Pete L. Clark Jan 26 '11 at 8:21

In some sense, the denseness of $\Bbb Q$ in $\Bbb R$ is implicit in the very same construction of $\Bbb R$.

There are a few equivalent ways to construct $\Bbb R$. One of them is to endow $\Bbb Q$ with a structure of metric space by defining the distance function $d(a,b)=|a-b|$ and define $\Bbb R$ to be the completion of the metric space $({\Bbb Q},d)$.

This is but an instance of a general construction: given any matric space $(X,d)$ one can construct its completion $\tilde X$ as the set of certain equivalence classes of Cauchy sequences in $X$. It turns out that one can endow $\tilde X$ with a natural metric space stucture such that: (1) $X$ embeds isometrically into $\tilde X$ and the image is a dense set, (2) $\tilde X$ is complete, and (3) if $\tilde Y$ is some metric space satisfying the just mentioned properties, then there is a canonical isometry of metric spaces between $\tilde X$ and $\tilde Y$ which is the identity on $X$.

Incidentally, it is possible to define distances in $\Bbb Q$ which are not equivalent to the one given by the usual absolute value (as above) but depend on the choice of a prime number $p$ and are defined arithmetically. The corresponding completions, the so-called fields of $p$-adic numbers ${\Bbb Q}_p$, are metric spaces not equivalent to $\Bbb R$ in which $\Bbb Q$ sits densely. A famous theorem by Ostrowski says that $\Bbb R$ and the ${\Bbb Q}_p$ are, up to equivalence, the only completions of $\Bbb Q$.


Here's a self contained proof. The Theorem numbers refer to Introduction to General Topology by Cain, but shouldn't affect comprehension.

By theorem 2.2, the $\mathbf{T}$-closure of a set $S$ consists of all points $p$ such that every member of $\mathbf{T}$ containing $p$ meets $S$. Symbollically, $cl(S)=\{p\in X:(\forall T\in\mathbf{T}:p\in\mathbf{T}),T\cap S\neq\emptyset\}$. Thus to show $cl(\mathbb{Q})=\mathbb{R}$ , we must show

$\mathbb{R}=\{p\in\mathbb{R}:(\forall T\in\mathbf{T}:p\in\mathbf{T}),T\cap\mathbb{Q}\neq\emptyset\}$

That is, we must show $\forall p\in\mathbb{R}$ , all members of $\mathbf{T}$ containing $p$ meet $\mathbb{Q}$. Since this is the usual topology, ie that generated by the usual pseudometric, a set is a member of $\mathbf{T}$ iff it is $d$-open. We know all $d$-open sets are unions of cells $(1.12)$. Thus showing each cell containing $p$ meets $\mathbb{Q}$ is sufficient to show each member of $\mathbf{T}$ containing $p$ meets $\mathbb{Q}$.

Let $p\in\mathbb{R}$ , and $B(p,\epsilon)$ . We must find some rational $q\in B(p,\epsilon)$ , ie $(p-\epsilon)<q<(p+\epsilon)$ . We can do so using the "Density Theorem" 2.4.8 in Real Analysis, by Bartle, 4th edition p. 44. This states: If $x,y$ are any real numbers with $x<y$ , then there exists a rational number $r\in\mathbb{Q}$ such that $x<r<y$ . Therefore, making the substitution $x=p-\epsilon,y=p+\epsilon$ , we have the desired result. Thus $\forall p\in \mathbb{R}$ every ball containing $p$ meets $\mathbb{Q}$. Thus $cl(\mathbb{Q)=R}$. $\square$

If anyone is interested, Bartle's proof of the "Density Theorem" is as follows:

"It is no loss of generality to assue $x>0$. Since $y-x>0$, it follows from Corollary 2.4.5 that there exists $n\in \mathbb{N}$ such that $1/n<y-x$. Therefore, we have $nx+1<ny$. If we apply Corollary 2.4.6 to $nx>0$, we obtain $m\in \mathbb{N}$ with $m-1\leq nx<m$. Therefore, $m\leq nx+1<ny$, whence $nx<m<ny$. Thus, the rational number $r:=m/n$ satisfies $x<r<y$."


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