Equivalent Cauchy Sequences have Same Modulus Limit

Theorem

Let $\struct {R, \norm { \, \cdot \, } }$ be a normed division ring.

Let $\sequence {x_n}$ and $\sequence {x_n}$ be Cauchy sequences in $R$.

Let $\displaystyle \lim_{n \to \infty} {x_n - y_n} = 0$

Then:

$\displaystyle \lim_{n \to \infty} \norm {x_n} = \lim_{n \to \infty} \norm {y_n}$

Proof

By Norm Sequence of Cauchy Sequence has Limit then $\displaystyle \lim_{n \to \infty} \norm {x_n}$ and $\displaystyle \lim_{n \to \infty} \norm {y_n}$ exist.

Let $l = \displaystyle \lim_{n \to \infty} \norm {x_n}$ and $m = \displaystyle \lim_{n \to \infty} \norm {y_n}$ then:

 $\displaystyle \lim_{n \to \infty } \paren { \norm {x_n } - \norm {y_n } }$ $=$ $\displaystyle l - m$ $\quad$ Difference Rule for real convergent sequencse $\quad$ $\displaystyle \implies \ \$ $\displaystyle \lim_{n \to \infty} \big\lvert \norm{x_n } - \norm{y_n } \big\rvert$ $=$ $\displaystyle \lvert l - m \rvert$ $\quad$ Modulus of limit $\quad$

By reverse triangle inequality, for $n \in \N$ then:

 $\displaystyle \big\lvert \norm{x_n } - \norm{y_n } \big\rvert$ $\le$ $\displaystyle \norm {x_n - y_n }$ $\displaystyle \to 0$ $\quad$ as $n \to \infty$ $\quad$

By the Squeeze Theorem then:

$\displaystyle \lim_{n \to \infty} \paren { \big\lvert \norm{x_n } - \norm{y_n } \big\rvert } = 0$

So $\lvert l - m \rvert = 0$ and therefore $l = m$

$\blacksquare$