Product Rule for Sequence in Normed Algebra

Theorem
Let $\GF \in \set {\R, \C}$.

Let $\struct {A, \norm {\, \cdot \,} }$ be a normed algebra over $\GF$.

Let $\sequence {a_n}_{n \in \N}$ and $\sequence {b_n}_{n \in \N}$ be sequences in $A$ converging to $a$ and $b$ respectively.

Then:


 * $a_n b_n \to a b$

Proof
From Convergent Sequence in Normed Vector Space is Bounded, there exists $M > 0$ such that:


 * $\norm {a_n} \le M$ for each $n \in \N$.

We have for $n \in \N$:

So from Sequence in Normed Vector Space Convergent to Limit iff Norm of Sequence minus Limit is Null Sequence, we have:


 * $a_n b_n \to a b$