How does $(a,b)$ where $a,b\in \mathbb{Q}$ form a basis for $\mathbb{R}$?

Question

Well I can't think how the open set $(\sqrt{2},\sqrt{3})$ can be formed from this basis.

Answer 1

The set $$A=\bigcup_{\begin{align}a,b\in&\Bbb Q\\\sqrt 2

Answer 2

Consider the decimeal representation of $\sqrt{2}$: $$\sqrt{2} = 1.414213562373 \dots$$ Let's make a rational sequence $a_n$ converging to $\sqrt{2}$, with $a_n \geq \sqrt{2}$ for all $n$. Take $a_1 = 1.5$, $a_2 = 1.42$, $a_3 = 1.415$, $a_4 = 1.4143$, etc: truncate the decimal representation of $\sqrt{2}$ and round it up. Clearly, $a_n \to \sqrt{2}$.

Similarly, you make a rational sequence $b_n$ converging to $\sqrt{3}$, with $b_n \leq \sqrt{3}$ for all $n$, by truncating and rounding down.

You then have a sequence of intervals $(a_n,b_n) \subseteq (\sqrt{2},\sqrt{3})$ with $a_n \to \sqrt{2}$ and $b_n \to \sqrt{3}$. Hence $$(\sqrt{2},\sqrt{3}) = \bigcup_n \;(a_n,b_n).$$

Answer 3

Let $a_n$ be a decreasing sequence of rational numbers converging to $\sqrt2 = 1.41421356\ldots$, for instance $1.5, 1.42, 1.415, 1.4143,\ldots$. Let $b_n$ be an increasing sequence of rational numbers converging to $\sqrt3=1.7320508\ldots$, for instance $1.7, 1.73, 1.732,\ldots$. Then we have that $$ (\sqrt 2, \sqrt3) = \bigcup_{n\in \Bbb N}(a_n, b_n) $$ so $(\sqrt 2, \sqrt3)$ is a union of basis elements and therefore open.