Consider the following Proposition.
For each $n\in\mathbf{N}$, assume we are given a closed interval $I_n
= [a_n, b_n] = \{x \in \mathbf{R} : a_n\leq x\leq b_n\}$ Assume also that each $I_n$ contains $I_{n+1}$. Then, the resulting nested
sequence of closed intervals $$I_1 \supseteq I_2 \supseteq
I_3\supseteq I_4\supseteq \dots$$ has a nonempty intersection, that is,
$\bigcap_{n=1}^{\infty}I_n\neq \varnothing$.
My Question how does the above property imply at the very least informally that there are no gaps in the real line?
Best Answer
Because it's false in $\Bbb Q$. For instance, if $a_n$ and $b_n$ are two sequences of rational numbers satisfying the "nested interval" condition and whose common limit is $\sqrt{2}$, then this intersection will be empty if you take intervals of rational numbers. Indeed, in $\Bbb R$, this intersection is $\{\sqrt{2}\}$.