Equivalent Cauchy Sequences have Equal Limits of Norm Sequences

Theorem

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

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

Let $\ds \lim_{n \mathop \to \infty} {x_n - y_n} = 0$.

Then:

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

Proof

Let:

$l = \ds \lim_{n \mathop \to \infty} \norm {x_n}$

and:

$m = \ds \lim_{n \mathop \to \infty} \norm {y_n}$

By Norm Sequence of Cauchy Sequence has Limit, both of these limits exist.

Then:

\(\ds \lim_{n \mathop \to \infty } \paren {\norm {x_n} - \norm {y_n} }\)

\(=\)

\(\ds l - m\)

Difference Rule for Real Sequences

\(\ds \leadsto \ \ \)

\(\ds \lim_{n \mathop \to \infty} \size {\norm {x_n} - \norm {y_n} }\)

\(=\)

\(\ds \size {l - m}\)

Modulus of Limit

By Reverse Triangle Inequality on Normed Division Ring, for $n \in \N$:

$l = \size {\norm {x_n} - \norm {y_n} } \le \norm {x_n - y_n} = \to 0$

as $n \to \infty$.

By the Squeeze Theorem:

$\ds \lim_{n \mathop \to \infty} \paren {\size {\norm {x_n} - \norm {y_n} } } = 0$

So:

$\size {l - m} = 0$

and therefore: $l = m$

$\blacksquare$