Take $\mathbb{R}$ in the standard (order) topology. Show that the closure of $\mathbb{Q}$ is $\mathbb{R}$.

I'm new to topology, self-studying using the Munkres book. In fact I'm new to proofs. In the book he has just defined a closed set, and the notion of a closure of a set. Here I attempt to do this proof. I find it difficult not to appeal to things that seem 'obvious' in the course of a proof (I flag a couple of these points below). But sometimes 'obvious' things are hard to prove indeed! How can I improve my proof? How would you prove it using only relatively elementary facts?

Let us consider some irrational $x \in \mathbb{Q}'$. For convenience, take $x>0$. I intend to show that for any such $x$, there's no neighborhood of $x$ that does not intersect $\mathbb{Q}$, so all such $x$ have to be in the closure in question, $\bar{\mathbb{Q}}$

A minimal open set $U$ that contains $x$ is $(r_1, r_2)$ with $r_1<x<r_2$ and $r_1,r_2\in \mathbb{R}$. But I shall show that no matter what we take for $r_1$ and $r_2$, $$r_1<r_2 \iff \exists q \in \mathbb{Q}: r_1<q<r_2$$

To begin, note $r_1 < r_2 \iff 0<r_2-r_1$.

Let us define $\Delta r \equiv r_2 - r_1$

Consider $(\Delta r)^{-1}$ -- it could be large if $\Delta r$ is small, but certainly$^1$ $\exists z \in \mathbb{Z}_+:z>(\Delta r)^{-1}$ and this implies that ${1 \over z}<\Delta r$. By definition, ${1 \over z} \in \mathbb{Q}$.

Next consider $m {1 \over z}$ for $m \in \mathbb{Z}$. If we take $m=0$, then $m {1\over z}<r_1$ (Otherwise we get $r_1 < 0 < r_2,$ and $0 \in \mathbb{Q}$). Also, if $m$ is large enough, then $m {1 \over z} > r_1$. Let $M=\{0\} \cup \{m \in \mathbb{Z}_+:m {1 \over z}<r_1\}$. Clearly $M$ is bounded by $zr_1$. Take $n$ to be the largest element$^2$ of $M$. Since $n \in M$, $n{1 \over z} < r_1$. Also, $(n+1){1 \over z}>r_1$, since if not, $(n+1) \in M, (n+1)>n$ which contradicts that n is the largest element in $M$.

Now $$n{1 \over z} < r_1 \wedge {1 \over z} < r_2 - r_1 \implies n{1\over z} + {1 \over z} < r_1 + (r_2 - r_1) \\ \implies (n+1){1 \over z} < r_2$$

Therefore, we have that $$ r_1 < {n+1 \over z} < r_2$$

But of course ${n + 1 \over z} \in \mathbb{Q}$ so this neighborhood of $x$ intersects $\mathbb{Q}$ no matter what interval $(r_1, r_2)$ containing $x$ we take, assuming $x$ is positive and irrational. This means that all neighborhoods of positive irrational numbers intersect $\mathbb{Q}$. I believe the argument for negative irrational numbers would go the same way without much different. This would lead to all neighborhoods of positive and negative irrational numbers intersecting $\mathbb{Q}$, thus putting them in $\bar{\mathbb{Q}}$. But the set of all positive or negative irrationals plus all rationals is itself $\mathbb{R}$ so we have $\bar{\mathbb{Q}} = \mathbb{R}$.

How did I do?

$^1$ I didn't show that $\forall r \in \mathbb{R}: \exists q \in \mathbb{Z}_+:q>r$. I find it difficult to prove something that seems so obvious. But I would like to see it proved once. How can we get that?

$^2$ I didn't show that $M$ has a largest element, how might I do that?

  • $\begingroup$ Haven't gone through it with a fine comb, but looks promising so far. Congrats on being scrupulous with "obvious" facts. Regarding the first, it's the Archimedian property, how to prove that depends on how $\mathbb{R}$ was defined/constructed. $\endgroup$ Commented Aug 31, 2013 at 22:12
  • $\begingroup$ Regarding the second, there is, by Archimedes, a $k \in \mathbb{Z}^+$ such that $k > z\cdot r_1$. Thus $M$ is finite, and a finite subset of $\mathbb{Z}$ (any ordered set) is empty or contains a largest element. $\endgroup$ Commented Aug 31, 2013 at 22:17
  • $\begingroup$ @DanielFischer thanks, I'll look up the archimedian property. As for the second point though, I see why $k>z\cdot r$ implies that $M$ is bounded from above in $\Bbb{R}$. But how do we get from that to the fact that it is finite? I know its obvious but understand how it is, to a beginner, quite shocking to realize that, say, $(0,1)\in \Bbb{R}$ is without a largest or smallest value. $\endgroup$ Commented Aug 31, 2013 at 23:02
  • $\begingroup$ It's a fundamental property of the natural numbers, for each $k \in \mathbb{N}$, the set $\{m \in \mathbb{N} : m < k\}$ is finite, of cardinality $k$. If you want a proof of that (and that's reasonable for sure), get a book on (elementary) set theory, where such things are usually proven directly from the axioms and definitions. It would take more space than available in the comments here to give what would resemble a complete proof of that. $\endgroup$ Commented Aug 31, 2013 at 23:11
  • $\begingroup$ Thanks, I understand. Actually, in the book I'm reading they define a set to be 'finite of cardinality k' if there's a bijection with $\{m \in \Bbb{N}: m < k \}, \quad k \in \Bbb{N}$, just like you say. So I'm satisfied (although in my case I have $k \in \Bbb R$). Thanks again. $\endgroup$ Commented Aug 31, 2013 at 23:28

2 Answers 2


Let $x<y$. Then choose $n$ such that $\frac{1}{n} < y-x$. Now note that we must have $\{\frac{m}{n}\}_m \cap (x,y) \neq \emptyset$. Hence any non-empty open set contains a rational.

Now consider $\overline{Q}$, which is closed by definition. Hence $(\overline{Q})^C$ is open. However, it cannot contain a rational, so it must be empty.

Addendum: To show that $\{\frac{m}{n}\}_m \cap (x,y) \neq \emptyset$, note that $\mathbb{R} = \cup_m [\frac{m}{n}, \frac{m+1}{n})$, and that the union is disjoint. Hence for some $m$, we have $[\frac{m}{n}, \frac{m+1}{n}) \cap (x,y) \neq\emptyset$. If $\frac{m}{n} \in (x,y)$ we are finished, so suppose $\frac{m}{n} \not\in (x,y)$. Then we must have $\frac{m}{n} \leq x <\frac{m+1}{n}$ (otherwise there is no overlap). Since $\frac{1}{n} < y-x$, we have $\frac{m}{n} + \frac{1}{n} = \frac{m+1}{n} < y$ and so $\frac{m+1}{n} \in (x,y)$.

  • $\begingroup$ Thanks @coppper. I guess you are taking it to be obvious that ${m \over n}_m \cap (x,y) \neq \emptyset$, which is fair enough, but could you show it formally? I know we don't always go to such extents, but I'd like to see it once. $\endgroup$ Commented Aug 31, 2013 at 22:35
  • 1
    $\begingroup$ I added an elaboration, hopefully this clarifies. $\endgroup$
    – copper.hat
    Commented Aug 31, 2013 at 22:56
  • $\begingroup$ Yes, that does help! $\endgroup$ Commented Aug 31, 2013 at 23:20
  • 1
    $\begingroup$ Question about the existence of ${1 \over n}:{1 \over n} < y - x$: I read a little just now about the Archimedian property, and see that it ensures such an n exists. So, do we normally have to take that as an axiom or part of the definition of $\Bbb{R}$? If we have sort of 'basic number theoretic axioms', for instance, about the closure of $\Bbb{R}$ under + and - etc, can we prove the Archimedian property for $\Bbb{R}$? $\endgroup$ Commented Aug 31, 2013 at 23:20
  • $\begingroup$ @EdwardNewell The AP is the statement that $\Bbb N$ is not bounded above. $\endgroup$
    – Pedro
    Commented Aug 31, 2013 at 23:29

It would be easier to do this. Pick $x,y$ reals. Assume $x>y$, so $x-y>0$. Archimedes says there is $n$ such that $n(x-y)=nx-ny>1$. Can you find an integer between $nx$ and $ny$ ${}^{1}$? Observe that $nx-ny>1$, which is important! If so, can you find a rational between $x$ and $y$? Why does this tell you that $\overline{ \Bbb Q}=\Bbb R$?

$1$. Spoiler Consider $m=\lfloor ny\rfloor+1$. Then by definition

$$ny<m\leq ny+1<ny+nx-ny=nx$$

DEF Let $x$ be a real number. We define $\langle x\rangle$ to be the greatest element of $$S=\{m\in\Bbb Z:m\leqslant x\}$$

This set is nonempty by the Archimedean property. In particular, we can reduce it to a finite set by choosing $m'$ such that $m'<x-1$. Then $m'\notin S$, and we may look at $$S=\{m\in\Bbb Z:m'< m\leqslant x\}$$

This is a non-empty finite set of integers, thus it admits a maximum (unique) element.

  • $\begingroup$ yes, for sure $\lfloor ny \rfloor + 1$ is between $nx$ and $ny$! But I don't have any definition for $\lfloor \cdot \rfloor$ yet. Can we do it from number theoretic axioms like commutativity and associativity of addition/multiplication, distributivity of multiplication over addition, the specially defined 0 and 1 and well-ordering? $\endgroup$ Commented Aug 31, 2013 at 22:53
  • $\begingroup$ I know, it's maybe nit picking, I just need to see it once. $\endgroup$ Commented Aug 31, 2013 at 22:54
  • $\begingroup$ @EdwardNewell You define $\lfloor ny\rfloor$ as the greatest integer $m$ such that $m\leqslant ny$. $\endgroup$
    – Pedro
    Commented Aug 31, 2013 at 22:54

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