Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent

Theorem

Let $K$ be a topological field.

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

Let $x \in X$.

Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $X$ with:

$x_n \to x$

Let $\lambda \in K$.

Then $\sequence {\lambda x_n}_{n \in \N}$ is a convergent sequence in $X$ with:

$\lambda x_n \to \lambda x$.

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 $\lambda x_n \to {\mathbf 0}_X = \lambda x$ as $n \to \infty$.

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

Now take $\lambda \ne 0_K$.

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

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

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

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

Then, we have:

$\lambda x_n \in U$ for $n \ge N$.

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

$\blacksquare$