Open Neighborhood of Dilation of Point in Topological Vector Space contains Pointwise Scalar Multiplication of Open Neighborhood of Scalar with Open Neighborhood of Vector

Theorem

Let $K$ be a topological field.

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

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

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

Then there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ in $X$ such that:

$D V \subseteq U$

where:

$D V = \set {\mu y : \mu \in D, \, y \in V}$

Proof

Equip $K \times X$ with the product topology.

From Box Topology on Finite Product Space is Product Topology, this is precisely the box topology.

Define $m : K \times X \to X$ by:

$\map m {\mu, y} = \mu y$

for each $\tuple {\mu, y} \in K \times X$.

From the definition of a topological vector space, $m$ is continuous.

In particular, $m$ is continuous at $\tuple {\lambda, x}$.

Hence, there exists an open neighborhood $W$ of $\tuple {\lambda, x}$ such that:

$m \sqbrk W \subseteq U$

From Basis for Box Topology and the definition of a topology generated by a synthetic basis, there exists an open neighborhood $D$ of $\lambda$ in $K$ and an open neighborhood $V$ of $x$ in $X$ such that:

$D \times V \subseteq W$

Then we have, from Image of Subset under Mapping is Subset of Image:

$m \sqbrk {D \times V} \subseteq m \sqbrk W \subseteq U$

Since:

$m \sqbrk {D \times V} = D V$

we have:

$D V \subseteq U$

$\blacksquare$