Why is a constant sequence a Cauchy sequence?

Question

I think that a constant sequence is a Cauchy sequence because of the definition of Cauchy sequence:

The sequence $\{q_n\}$ of rationals is a Cauchy sequence if, for every $\epsilon>0$, where $\epsilon$ is a rational number, there is a natural number (with the $0$) $N$ such that: $|q_i-q_j|<\epsilon$ for all $i,j\geq N$

For any $\epsilon$ I choose, let's say $0.00001$, there will always be two elements of the sequence such that distance $|q_i-q_j|<\epsilon=0.00001$ (where $i,j\geq N$) because since all the elements of this sequence are all equal, this distance is always $0$. Is this correct? This is kinda even why every convergent sequence is a Cauchy sequence right?

Thank you!

Answer 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.

Answer 2

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

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