Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide

Theorem

Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $M_a$.

Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, that is:

$\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$

Then $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$.

Proof

We have that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $M_a$.

Then:

$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$

By equivalence of norms:

$\exists M \in \R_{> 0} : \norm {x_n - x_m}_b \le M \norm {x_n - x_m}_a < M \epsilon_a$

Let $\epsilon_b := M \epsilon_a$

Then:

$\forall \epsilon_b \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n, m > N \implies \norm {x_n - x_m}_b < \epsilon_b$

Therefore, $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$.

$\blacksquare$

Sources

• 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces