Equivalent Cauchy Sequences have Equal Limits of Norm Sequences

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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:

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

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

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

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 $l = m$


$\blacksquare$


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