Prove that ℚ is disconnected as a subset of ℝ

Question

This is the proof I've come up with...

Using the following definition: Let S be a metric space. E ⊂ S is disconnected iff there are disjoint, non empty subsets A and B that are open in E and such that E = A ∪ B.

Let p be irrational, then ℚ can be written as (-∞,p) ∪ (p,∞) Here, A = (-∞,p) and B = (p,∞) , which is the union of two disjoint open sets that are nonempty. Therefore, ℚ is disconnected.

We also know that E ⊂ ℝ is connected iff if a, b ∈ E and if a < x < b, then x ∈ E. So, for ℚ to be disconnected in ℝ it can't be an interval.

ℚ is not an interval because it can be written as above: (-∞,p) ∪ (p,∞), where p is an irrational between two rationals a, b.

To me the second piece seems redundant, but I want to make sure I address the portion about ℚ as a subset of ℝ. Is there any way I can make this more concise?

Answer 1

To avoid repeating the direct proof (i.e. the first proof you gave using the definition of connectedness), you could say, for instance, that $\mathbb{Q}$ is not a real interval since $1 \in \mathbb{Q}$ and $2 \in \mathbb{Q}$ but $1 < \sqrt{2} < 2$ and $\sqrt{2} \not\in \mathbb{Q}$.

Answer 2

The second part is indeed redundant. The portion about $\mathbb{Q}$ as a subset of $\mathbb{R}$ has already been addressed when you used the fact that the open subsets of $\mathbb{Q}$ are the same as the intersections with $\mathbb{Q}$ of open subsets of $\mathbb{R}$, so for example $(-\infty,p) \cap \mathbb{Q}$ is open in $\mathbb{Q}$.