Convergence in Direct Product Norm

Theorem

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

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

Let $\norm {\, \cdot \,}_{X \times Y}$ be the direct product norm.

Let $\sequence {\tuple {x_n, y_n} }_{n \mathop \in \N}$ be a sequence in $X \times Y$.

Then:

$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$

if and only if:

$x_n$ converges to $x$ in $X$

and:

$y_n$ converges to $y$ in $Y$.

Proof

Neccessary Condition

Suppose that:

$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$

Then, for each $\epsilon > 0$, there exists $N \in \N$ such that:

$\ds \norm {\tuple {x_n, y_n} - \tuple {x, y} }_{X \times Y} < \epsilon$

for all $n \ge N$.

This is equivalent to:

$\ds \norm {\tuple {x_n - x, y_n - y} }_{X \times Y} < \epsilon$

by the definition of induced component-wise operations.

From the definition of the direct product norm, this is equivalent to:

$\ds \map \max {\norm {x_n - x}_X, \norm {y_n - y}_Y} < \epsilon$

This is equivalent to:

$\norm {x_n - x}_X < \epsilon$

and:

$\norm {y_n - y}_Y < \epsilon$

for $n \ge N$.

That is:

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

$\Box$

Sufficient Condition

Suppose that:

$x_n$ converges to $x$ in $X$

and:

$y_n$ converges to $y$ in $Y$.

Let $\epsilon > 0$.

Then we can pick $N_1 \in \N$ such that:

$\norm {x_n - x}_X < \epsilon$

for $n \ge N_1$.

Similarly, we can pick $N_2 \in \N$ such that:

$\norm {y_n - y}_Y < \epsilon$

Let:

$N = \max \set {N_1, N_2}$

Then, for $n \ge N$, we have:

$\norm {x_n - x}_X < \epsilon$

and:

$\norm {y_n - y}_Y < \epsilon$

so that:

$\map \max {\norm {x_n - x}_X, \norm {y_n - y}_Y} < \epsilon$

From the definition of the direct product norm, we have:

$\norm {\tuple {x_n - x, y_n - y} }_{X \times Y} < \epsilon$

From the definition of induced component-wise operations, we have:

$\norm {\tuple {x_n, y_n} - \tuple {x, y} }_{X \times Y} < \epsilon$

for all $n \ge N$.

So:

$\tuple {x_n, y_n}$ converges to $\tuple {x, y}$ in $X \times Y$

$\blacksquare$