Scalar Multiple of Convergent Net in Topological Vector Space is Convergent

Theorem

Let $K$ be a topological field.

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

Let $\struct {\Lambda, \preceq}$ be a directed set.

Let $x \in X$ and $\mu \in K$.

Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net converging to $x$.

Then $\family {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$.

Proof

For ease of reading, let $\succeq$ be the inverse relation of $\preceq$.

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

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

From Constant Net is Convergent, we have that $\mu x_\lambda$ converges to ${\mathbf 0}_X$.

So $\sequence {\lambda x_n}_{n \in \N}$ converges to $\mu x$ for $\mu = 0_K$.

Now take $\mu \ne 0_K$.

Let $U$ be an open neighborhood of $\mu x$.

From Dilation of Open Set in Topological Vector Space is Open, $\mu^{-1} U$ is an open neighborhood of $x$.

Since $\sequence {x_n}_{n \in \N}$ converges, there exists $\lambda_0 \in \Lambda$ such that:

$x_n \in \mu^{-1} U$ for $\lambda \succeq \lambda_0$.

Then, we have:

$\mu x_n \in U$ for $\lambda \succeq \lambda_0$.

Since $U$ was an arbitrary open neighborhood of $\mu x$, we have that $\sequence {\mu x_\lambda}_{\lambda \in \Lambda}$ converges to $\mu x$ as required.

$\blacksquare$