# Disconnectedness of the rationals with the subspace topology

I have tried to prove that $\mathbb{Q}\subset \mathbb{R}$ equipped with the subspace topology is a disconnected space. I'd like to make sure my attempted proof is correct since topological properties of $\mathbb{Q}$ seem like nice things to be familiar with.

First, a base for the subspace topology on $\mathbb{Q}$ is $\{(a,b)\cap \mathbb{Q}\,|\,a<b \in \mathbb{R}\}$, yes?

To separate the rationals into two disjoint open sets (a space is disconnected if it can be written as the union of two nonempty disjoint open sets), we need the end of one 'interval' to lie between two rational numbers, which intuitively seems to be problematic at first, since we can find rational numbers arbitrarily close to each other.

However, the rationals are countable. So label them as $a_1, a_2, ...\,$. Can we then take two arbitrary rationals, $a_i$ and $a_{i+1}$, and then take the relatively open sets $(-\infty,a_i)$ and $(a_{i+1},\infty)$, which are clearly disjoint, and whose union is clearly $\mathbb{Q}$ as a separation of $\mathbb{Q}$?

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The issue with what you have is that you need there to be no rationals between $a_i$ and $a_{i+1}.$ (And that is clearly impossible). To fix this, you can do what William did, and use irrationals as boundaries. –  only Aug 30 '12 at 2:36

$\mathscr{B}=\{(a,b)\cap\Bbb Q:a,b\in\Bbb R\text{ and }a<b\}$ is indeed a base for the topology that $\Bbb Q$ inherits from the usual topology on $\Bbb R$. After that, though, you’ve gone badly astray. If $p$ and $q$ are distinct rational numbers with $p<q$, then the union of $(\leftarrow,p)\cap\Bbb Q$ and $(q,\to)\cap\Bbb Q$ is definitely not all of $\Bbb Q$: it misses every rational number in the interval $[p,q]$.

It’s true that $\Bbb Q$ is a countable set, but no matter how you index it as $\{q_n:n\in\Bbb N\}$, it won’t be true that $q_n$ and $q_{n+1}$ are adjacent: no two rational numbers are adjacent. If $p$ and $q$ are distinct rational numbers, then $\frac12(p+q)$ lies strictly between them and is also rational.

In order to get a simple disconnection of $\Bbb Q$, pick an irrational number $\alpha$; $\sqrt2$ will do nicely. Then let $U=(\leftarrow,\alpha)\cap\Bbb Q$ and $V=(\alpha,\to)\cap\Bbb Q$; $(\leftarrow,\alpha)\cup(\alpha,\to)=\Bbb R\setminus\{\alpha\}$, and $\alpha\notin\Bbb Q$, so $U\cup V=\Bbb Q$, and certainly $U\cap V=\varnothing$. Finally, if $p\in U$, then $p\in(p-1,\alpha)\subseteq U$, where $(p-1,\alpha)\in\mathscr{B}$, so $U$ is open in $\Bbb Q$. Similarly, if $p\in V$, $(\alpha,p+1)$ is a member of $\mathscr{B}$ containing $p$ and contained in $V$, so $V$ is also open in $\Bbb Q$.

You can do even more: for any two rational numbers $p$ and $q$, there is an irrational number $\alpha$ strictly between them, so you can find a separation of $\Bbb Q$ that puts $p$ and $q$ into opposite sides of the separation. Thus, $\Bbb Q$ is totally disconnected.

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How did I miss that? Clearly I should have thought about this a little longer. Thanks. –  Alex Petzke Aug 30 '12 at 3:46

Here is one way of showing this. Take an irrational like $\sqrt{2}$. $(-\infty, \sqrt{2})$ and $(\sqrt{2}, \infty)$ are both open in the topology of $\mathbb{R}$. Hence $U_1 = (-\infty, \sqrt{2}) \cap \mathbb{Q}$ and $U_2 = (\sqrt{2}, \infty) \cap \mathbb{Q}$ are both open in $\mathbb{Q}$ with the relative toplogy. Then $\mathbb{Q} = U_1 \cup U_2$. $\mathbb{Q}$ is the union of two disjoint open set. $\mathbb{Q}$ is not connected.

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