Let $\left\{ x_k \right\}$ be a convergent sequence in $R$. Show that the set $\left\{ x_1, x_2, \dots \right\}$ has zero content.
My book's definition of zero content is:
If a set $S \in \mathbb{R}$ has zero content, then for any $\epsilon > 0$, there is a finite collection of intervals $I_1, \dots , I_{L}$ such that (i) $S \subset \bigcup_{1}^{L} I_{l}$, and (ii) the sum of the lengths of the $I_{l}$'s is less then $\epsilon$.
Really stuck on this problem. I don't know how to prove this symbolically. Would appreciate any help!
Best Answer
Let $\lim x_k=x$. Then there exists some $N\in\mathbb N^*$ such that $|x_n-x|<\varepsilon/8$ for all $n\ge N$. Then you can choose $N$ intervals in the following way:
(i) $I_N=[x-\varepsilon/8, x+\varepsilon/8]$. The length of $I_N$ is $\varepsilon/4$;
(ii) For each $n=1,2,\dots, N-1$, let $I_n=[x_n-\frac{\varepsilon}{8(N-1)}, x_n+\frac{\varepsilon}{8(N-1)}]$. The length of each of these $N-1$ intervals is $\frac{\varepsilon}{4(N-1)}$.
Then it is clear that $S\subseteq\bigcup_{n=1}^N I_n$, and the sum of the lengths is $\varepsilon/2$.