Space of Zero-Limit Sequences with Supremum Norm forms Banach Space

Theorem
Let $c_0$ be the space of zero-limit sequences.

Let $\norm {\, \cdot \,}_\infty$ be the supremum norm.

Then $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.

Proof
Let $\sequence {a_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.

Let $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.

By Space of Zero-Limit Sequences with Supremum Norm forms Normed Vector Space, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a subspace of $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.

Hence, $\sequence {a_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$.

By Space of Bounded Sequences with Supremum Norm forms Banach Space, $\sequence {a_n}_{n \mathop \in \N}$ converges to $a \in \ell^\infty$.

Denote $a_n = \sequence {a_n^{\paren m}}_{m \mathop \in \N}$ and $a = \sequence {a^{\paren m}}_{m \mathop \in \N}$.

By definition of convergent sequences:


 * $\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {a_n - a}_\infty < \epsilon$

Then:

Since $a_n \in c_0$:


 * $\forall \epsilon \in \R_{\mathop > 0} : \exists M \in \R_{\mathop > 0} : \forall m \in \N : m > M \implies \size {a_n^{\paren m}} < \epsilon$

For all $m > M$ we also have that:

In other words:


 * $\forall \epsilon' \in \R_{\mathop > 0} : \exists M' \in \R_{\mathop > 0} : \forall m \in \N : m > M' \implies \size {a^{\paren m}} < \epsilon'$

where $\epsilon' = 2\epsilon$ and $M' = M$.

By definition of zero-limit sequences, $a \in c_0$.

Therefore, in $\struct {c_0, \norm {\, \cdot \,}_\infty}$ a Cauchy sequence is also convergent in $\struct {c_0, \norm {\, \cdot \,}_\infty}$.

By definition, $\struct {c_0, \norm {\, \cdot \,}_\infty}$ is a Banach space.