Combination Theorem for Cauchy Sequences/Product Rule

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Theorem

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring.

Let $\sequence {x_n}$, $\sequence {y_n} $ be Cauchy sequences in $R$.

Then:

$\sequence {x_n y_n}$ is a Cauchy sequence.


Proof

Because $\sequence {x_n} $ is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.

Suppose $\norm {x_n} \le K_1$ for $n = 1, 2, 3, \ldots$.

Because $\sequence {y_n} $ is a is a Cauchy sequence, it is bounded by Cauchy Sequence is Bounded.

Suppose $\norm {y_n} \le K_2$ for $n = 1, 2, 3, \ldots$.

Let $K = \max \set {K_1, K_2}$.

Then both sequences are bounded by $K$.


Let $\epsilon > 0$ be given.

Then $\dfrac \epsilon {2K} > 0$.

Since $\sequence {x_n}$ is a Cauchy sequence, we can find $N_1$ such that:

$\forall n, m > N_1: \norm {x_n - x_m} < \dfrac \epsilon {2 K}$

Similarly, $\sequence {y_n} $ is a Cauchy sequence, we can find $N_2$ such that:

$\forall n, m > N_2: \norm {y_n - y_m} < \dfrac \epsilon {2 K}$

Now let $N = \max \set {N_1, N_2}$.

Then if $n, m > N$, both the above inequalities will be true.

Thus $\forall n, m > N$:

\(\ds \norm {x_n y_n - x_m y_m}\) \(=\) \(\ds \norm {x_n y_n - x_n y_m + x_n y_m - x_m y_m}\)
\(\ds \) \(\le\) \(\ds \norm {x_n y_n - x_n y_m} + \norm {x_n y_m - x_m y_m}\) Norm Axiom $\text N 3$: Triangle Inequality
\(\ds \) \(=\) \(\ds \norm {x_n \paren {y_n - y_m } } + \norm {\paren {x_n - x_m} y_m}\)
\(\ds \) \(=\) \(\ds \norm {x_n} \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot \norm {y_m}\) Norm Axiom $\text N 2$: Multiplicativity
\(\ds \) \(\le\) \(\ds K \cdot \norm {y_n - y_m} + \norm {x_n - x_m} \cdot K\) as both sequences are bounded by $K$
\(\ds \) \(\le\) \(\ds K \cdot \dfrac \epsilon {2K} + \dfrac \epsilon {2K} \cdot K\)
\(\ds \) \(=\) \(\ds \dfrac \epsilon 2 + \dfrac \epsilon 2\)
\(\ds \) \(=\) \(\ds \epsilon\)

Hence:

$\sequence {x_n y_n}$ is a Cauchy sequence in $R$.

$\blacksquare$


Sources

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