Direct Product of Banach Spaces is Banach Space

Theorem
Let $\struct {X, \norm {\, \cdot \,}}$ and $\struct {Y, \norm {\, \cdot \,}}$ be normed vector spaces.

Let $V = X \times Y$ be a direct product of vector spaces $X$ and $Y$ together with induced component-wise operations.

Let $\norm {\tuple {x, y} }$ be the direct product norm.

Suppose $X$ and $Y$ are Banach spaces.

Then $V$ is a Banach space.

Proof
Let $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ be a Cauchy sequence in $V$:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n > N: \norm {\tuple {x_n, y_n} - \tuple {x_m, y_m} } < \epsilon$

We have that:

Hence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy sequence in $X$.

By assumption, $X$ is a Banach space.

Therefore, $\sequence {x_n}_{n \mathop \in \N}$ converges to $x \in X$.

By analogous arguments, $\sequence {y_n}_{n \mathop \in \N}$ converges to $y \in Y$.

Then there exists $\epsilon \in \R_{> 0}$ such that:


 * $\exists N_x \in \N : \forall n \in \N : n > N_x : \norm {x - x_n}_n < \epsilon$


 * $\exists N_y \in \N : \forall n \in \N : n > N_y : \norm {y - y_n}_n < \epsilon$

Let $N := \map \max {N_x, N_y}$.

Then:


 * $\forall n \in \N : n > N : \paren {\norm {x - x_n}_n < \epsilon} \land \paren{\norm {y - y_n}_n < \epsilon}$

Thus:

Therefore, $\sequence {\tuple {x_n, y_n}}_{n \mathop \in \N}$ converges to $\tuple {x, y}$.

By definition, $V$ is a Banach space.