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Let $\mathbb{Q}$ be the set of rationals with its usual topology based on distance: $$d(x,y) = |x-y|$$ Suppose we can only use axioms about $\mathbb{Q}$ (and no axiom about $\mathbb{R}$, the set of reals). Then how can we show that $\mathbb{Q}$ is topologically disconnected, i.e.: there exist two open sets $X$ and $Y$ whose union is $\mathbb{Q}$?

If we were allowed to use axioms about $\mathbb{R}$, then we could show that for any irrational number $a$:

  • if $M$ is the intersection of $]-\infty, a[$ with the rationals, then $M$ is an open set of $\mathbb{Q}$
  • if N is the intersection of $]a, +\infty[$ with the rationals, then $N$ is an open set of $\mathbb{Q}$
  • $\mathbb{Q}$ is the union of $M$ and $N$. CQFD.

But if we are not allowed to use axioms about $\mathbb{R}$, just axioms about $\mathbb{Q}$?

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For some basic information about writing math at this site see e.g. here, here, here and here. – Julian Kuelshammer Jan 18 '13 at 13:25

The rationals is the union of two disjoint open sets $\{x\in\mathbb{Q}:x^2>2\}$ and $\{x\in\mathbb{Q}:x^2<2\}$.

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Thanks. But then we need to show, using only the axioms of Q, that these are indeed open sets in Q. This seems equivalent to showing that the square root of 2 is not rational. I suppose the Greeks already had a proof of that? IF so, how did they show it? (using only axioms about Q) – user58762 Jan 18 '13 at 13:18
@David: Suppose $2 = (p/q)^2$ with $p$ and $q$ integers with $\gcd(p, q) = 1$. Then $2q^2 = p^2$ $\Rightarrow$ $2$ divides $p^2$ $\Rightarrow$ $2$ divides $p$ $\Rightarrow$ $4$ divides $p^2$ $\Rightarrow$ $4$ divides $2q^2$ $\Rightarrow$ $2$ divides $q^2$ $\Rightarrow$ $2$ divides $q$ $\Rightarrow$ contradiction with $\gcd(p, q) = 1$. There's no mention of the real numbers. – Tunococ Jan 18 '13 at 13:22
@David: I posted a proof here that any algebraic integer which is rational, is an integer. Since $\sqrt2$ satisfies $x^2-2=0$, it is an algebraic integer. Since there is no integer so that $x^2=2$, $\sqrt2$ cannot be rational. – robjohn Jan 18 '13 at 13:23
@Tunococ: However, you have to be able to for any rational $p$ with $p^2 > 2$ there exists a neighborhood $V(p)$ so that if $q \in V(p)$ then $q^2 > 2$ as well. And similarly if $p^2 < 2$. That $\{x \in \mathbb{Q} : x^2 >2$ is open since it's the intersection of $(\sqrt{2},\infty) \cap \mathbb{Q}$ only follows by considering $\mathbb{Q}$ as a subspace of $\mathbb{R}$, something the OP apparently wants to avoid. – anonymous Jan 18 '13 at 13:33
@anonymous: $x^2$ is a continuous function and the inverse image of open sets is open. Thus, $\{x\in\mathbb{Q}:x^2>2\}$ and $\{x\in\mathbb{Q}:x^2<2\}$ are open. Since $x^2$ cannot be both $>2$ and $<2$, they are disjoint. The only thing left to show is that their union is $\mathbb{Q}$. That follows as soon as we show there is no rational so that $x^2=2$. – robjohn Jan 18 '13 at 13:38

$(\mathbb{Q},d)$ is metric space. Since connceted metric space having more that more than one point is uncountable and $\mathbb{Q}$ is countable, $(\mathbb{Q},d)$ must be disconnected.

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Note the argument there uses the properties of $\Bbb R$. – David Mitra Jan 18 '13 at 14:15

A la Cantor's first proof of uncountability of $\mathbb R$:

Fix an enumeration of the countable set $\mathbb Q$. In the following, expressions like "first rational" refer to this enumeration. Let $a_0$ be the first rational. Let $b_0$ be the first rational bigger than $a_0$. For $n=0,1,2\ldots$ do the following: Given rationals $a_n<b_n$ let $c,d$ be the first two rationals between $a_n$ and $b_n$ (such numbers exist, e.g. $a_n<\frac{2a_n+b_n}3<\frac{a_n+2b_n}3<b_n$). Let $a_{n+1}=\min\{c,d\}$, $b_{n+1}=\max\{c,d\}$. Note that this implies $$\tag1a_n<a_{n+1}<b_{n+1}<b_n.$$

Let $$\tag2U=\{x\in\mathbb Q\mid \exists n\colon x<a_n\}, \quad V=\{x\in\mathbb Q\mid \exists n\colon x>b_n\}.$$ Then $U,V$ are nonempty open subsets of $\mathbb Q$ with $U\cap V=\emptyset$, $U\cup V=\mathbb Q$:

  • nonempty: $a_0-1\in U$, $b_0+1\in V$
  • open: $x\in U$ implies $x<a_n$ for some $n$, implies $y\in U$ for all $y$ with $|y-x|<a_n-x$. Similarly for $V$
  • disjoint: Assume $x\in U\cap V$. Then $b_n<x<a_m$ for some $n,m$. Repeated application of (1) produces $b_{\max\{n,m\}}\le\ldots \le b_n<a_m\le\ldots\le a_{\max\{n,m\}}$, contradiction
  • covering: If $x\in \mathbb Q$, then there is an index $m$ at which it occurs in the enumeration of $\mathbb Q$. If $x$ is not among the first $2n$ numbers $a_0,\ldots a_{n-1}, b_0,\ldots,b_{n-1}$ selected, then either $m>2n$ or $x$ is not between $a_{n-1}$ and $b_{n-1}$. The first option is ruled out for $n$ big enough, the second implies $x<a_n$ or $x>b_n$, hence $x\in U\cup V$.
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Thanks for your answer. This gives another proof of the desired result (Q is topologically disconnected). How is that proof related to the uncountability of R? – user58762 Jan 18 '13 at 16:28

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