Why is a constant sequence a Cauchy sequence?

Solution 1:

To show a sequence $(x_n)$ is Cauchy, you don't fix $\epsilon.$ You want to show for every $\epsilon>0,$ there exists $N \in \mathbb{N}$ such that $|x_n-x_m|<\epsilon$ for all $n,m \geq N.$

This definition is for any sequence $(x_n)$ in $\mathbb{R}.$ The sequence need not consist of only rationals.

Coming to a constant sequence, let $x_n = c \in \mathbb{R}$ for all $n \in \mathbb{N}.$ To show $(x_n)$ is Cauchy, let $\epsilon>0$ be arbitrary and $N=1.$ Then $$|x_n-x_m|=|c-c|=0<\epsilon\; \forall n,m\geq N.$$ Hence $(x_n)$ is Cauchy.

Solution 2:

Short answer: The distance between any two elements is zero, which is less than any chosen $\epsilon > 0$.

I.e. $\forall e>0, \exists N' \in N, \forall n,m \ge N' :\ |x_n - x_m| = 0 < e$