Scalar Multiple of Cauchy Sequence in Topological Vector Space is Cauchy Sequence

Theorem

Let $K$ be a topological field.

Let $X$ be a topological vector space over $K$.

Let $\sequence {x_n}_{n \in \N}$ be a Cauchy sequence in $X$.

Let $\lambda \in K$.

Then $\sequence {\lambda x_n}_{n \in \N}$ is a Cauchy sequence in $X$.

Proof

First consider the case $\lambda = 0_K$.

Then $\lambda x_n = {\mathbf 0}_X$ for each $n \in \N$.

From Constant Sequence in Topological Space Converges, we have that $\lambda x_n \to {\mathbf 0}_X$ as $n \to \infty$.

From Convergent Sequence in Topological Vector Space is Cauchy, $\sequence {\lambda x_n}_{n \in \N}$ is therefore Cauchy.

Now take $\lambda \ne 0$.

Let $U$ be an open neighborhood of ${\mathbf 0}_X$.

From Dilation of Open Set in Topological Vector Space is Open, $\lambda^{-1} U$ is an open neighborhood of ${\mathbf 0}_X$.

Since $\sequence {x_n}_{n \in \N}$ is Cauchy, there exists $N \in \N$ such that:

$x_n - x_m \in \lambda^{-1} U$ for $n, m \ge N$.

We therefore have:

$\lambda x_n - \lambda x_m \in U$ for $n, m \ge N$.

Since $U$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $\sequence {\lambda x_n}_{n \in \N}$ is Cauchy.

$\blacksquare$