Every sequence of real numbers can be covered by a sequence of open intervals of arbitrarily small length.

Question

How do I see that every sequence of real numbers can be covered by a sequence of open intervals of arbitrarily small length, namely that if $x_1, x_2, x_3, \dots$ is a sequence of real numbers and $\epsilon$ is positive, then there exists a sequence of open intervals $(a_n, b_n)$ with $a_n < x_n < b_n$ such that$$\sum_{n=1}^\infty (b_n - a_n) < \epsilon?$$

Answer 1

Let $(a_n,b_n) = (x_n - 2^{-n - 2} \epsilon, x_n + 2^{-n - 2} \epsilon)$.

Then $x_n \in (a_n,b_n)$ and $\sum b_n - a_n < \epsilon$.

Answer 2

Because every sequence is a function from $\mathbb N \to \mathbb R$ so in other words you want to find the measure of $\mathbb N$ in $\mathbb R$ and every countable set has measure $0$ i.e $\mathbb N$ can be easily covered by open balls of radius $\epsilon $ by $\mathbb N=\cup _{n\in \mathbb n}(n-\epsilon,n+\epsilon)$ for each $\epsilon>0$