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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?

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    $\begingroup$ It depends on what you define as gap. In fact, one can extend $\Bbb R$ to so-called hyperreals by adding infinitesimal elements that are in "gaps" between reals. For example there are then infinitesimals $\epsilon$ such that $\epsilon>0$ and also $\epsilon<a$ for all real positive $a$ $\endgroup$ – Hagen von Eitzen Jul 7 '18 at 7:35
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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 $\{\sqrt{2}\}$.

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