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I was playing around with an exercise which shows that if $A \neq \varnothing \subseteq \mathbb{R}$ is an open set, then $A \cap \mathbb{Q} \neq \varnothing$. I then extended the problem to a countable union of nonempty open sets, and then intuitively saw that we can always find at least one distinct rational number for each of the sets, and then reasoned that there are infinitely many rational numbers. Naturally, tried to prove it, so here is my attempt.

If $\mathscr{A} = \bigcup_{k=1}^{\infty} A_k \subseteq \mathbb{R}$ is a countable union of nonempty open sets, allowing $x \in \mathscr{A}$ means that there exists a $k \in \mathbb{N}$ such that $x \in A_k$. Since $A_k$ is open, there exists a neighborhood $N_{\epsilon}(x)$ such that $(x-\epsilon,x+\epsilon) \subseteq A_k$. Then between $x-\epsilon$ and $x+\epsilon$, there exists a rational number $q$ between them, so that $q \in N_{\epsilon}(x)$, and therefore $q \in \mathscr{A}$.

Since we can always find distinct rational numbers for each set in $\mathscr{A}$, and there are infinitely many such sets, we will find infinitely many rational numbers, so $\mathbb{Q}$ is infinite.

Any thoughts or concerns? Thanks in advance.

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  • $\begingroup$ How do you know that the rational numbers you find for each $A_k$ are distinct? If $A_k$ is the open interval $(-k,k)$ then I can just take $0$ for each of them. Also: you seem to assume that your union is infinite, which means you already know that there are infinitely many integers $k$ from $1$ to $\infty$....as all those integers are, of course, rational then you are assuming what you intend to prove. $\endgroup$ – lulu May 19 '16 at 21:54
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    $\begingroup$ Well, my concern is circularity. 1) How do you know there exist a countable union of (distinct) nonempty ope sets in the first place. Also "we can always find distinct rational numbers for each set in A" ... how do you know the rational numbers are distinct? (Although that's probably easily resolved. Bu there existing a countable union of nonempty open sets isn't.) $\endgroup$ – fleablood May 19 '16 at 21:56
  • $\begingroup$ You could choose 0 for each $A_k$, but I'm saying that it is always possible to find a distinct rational number. But I do accept your second point, so my reasoning is circular. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 21:57
  • $\begingroup$ @fleablood Yeah, the circularity is the real issue. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 21:58
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    $\begingroup$ " but I'm saying that it is always possible to find a distinct rational number" How do you know? Well you can pick the rationals, call them $q_k$ and then find the neighborhood that is the min between epsilon and the minimum d(q_k, x) so far. But you can't declare definitively that an uncountable number of distinct open sets exist. You just can't. Not without first assuming the rationals or at least the reals are infinite. $\endgroup$ – fleablood May 19 '16 at 22:08
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This proof doesn't work; you need the additional hypothesis that the $A_k$ are pairwise disjoint. Otherwise, you can't know that you're not just picking the same $q$ for each set. For instance, if you had $$\mathscr A =\bigcup_{k=1}^{\infty}(0,1)$$ you would find that $\frac{1}2$ is a rational number everywhere in it.

Also, you barely use the fact that $\mathscr A$ is a countable union of other things in your proof. More or less, you find that it is open because it is a union of open sets, and then the rest of the proof proceeds using only that $\mathscr A$ is open.

What does work is proving that there are infinitely many rational numbers in a set like $$\mathscr A = \bigcup_{k=-\infty}^{\infty}(k,k+1)$$ since this is a countable union of disjoint sets, each of which contains a rational number. You can also note that, if $A$ is a non-empty open set and $q_1,q_2,\ldots,q_n$ is a finite sequence of rationals in it, then $A\setminus \{q_1,q_2,\ldots,q_n\}$ is a non-empty open set and thus contains a rational distinct from $q_1,q_2,\ldots,q_n$. One can therefore find that there are infinitely many rationals in every open set.

Of course, if we're really just after proving that the rational numbers are infinite, it would seem easier to note that the integers are infinite and are a subset of the rationals.

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  • $\begingroup$ Good answer. I was wondering, though, if I say that it is always possible to find a distinct rational number (and this process can be completed by repeatedly looking for a rational between a pair of distinct real number $x<y$ in a set $A_k$ until a distinct rational is found), would such a statement be valid? If not, I would like to know more. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 22:06
  • $\begingroup$ @BenedictVoltaire You certainly can show that you can always find a rational number distinct from one you've already chosen - if you want to stay very elementary, you need to show that there is some interval $(x,y)$ which contains no rational number you've yet used. You can also use tricks like "find a rational number in $\mathscr A$ smaller than the smallest we've yet chosen," though some justification is needed. More abstractly, the validity of this amounts to showing that finite sets are closed and not open. $\endgroup$ – Milo Brandt May 19 '16 at 22:13
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    $\begingroup$ As it stands, no, the statement isn't valid but as Milo Brandt point's out you can always find a distinct new rational in $q_{i+1} \in (x, q_i)$. Thus by induction the rationals have at least the same cardinality of an iductive set. But I have to be honest, but using the premise "the rationals are dense" as a mandate from heaven makes me nervous. I'm pretty sure in showing the rationals are dense in the first place, their infinite cardinality was established as an unavoidable by product. $\endgroup$ – fleablood May 19 '16 at 22:14
  • $\begingroup$ @MiloBrandt Very nice insight. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 22:16
  • $\begingroup$ @fleablood Nicely worded. $\endgroup$ – Jeremiah Dunivin May 19 '16 at 22:16

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