Dilation of Closed Set in Topological Vector Space is Closed Set

Theorem

Let $K$ be a topological field.

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

Let $F$ be a closed set in $X$.

Let $\lambda \in K \setminus \set {0_K}$.

Then $\lambda F$ is a closed set in $X$.

Proof 1

We aim to show that $X \setminus \paren {\lambda F}$ is open.

Since $F$ is closed, $X \setminus F$ is open.

It follows from Dilation of Open Set in Topological Vector Space is Open that $\lambda \paren {X \setminus F}$ is open.

From Dilation of Complement of Set in Vector Space, we have:

$X \setminus \paren {\lambda F} = \lambda \paren {X \setminus F}$.

Since we have established that $\lambda \paren {X \setminus F}$ is open, it follows that $X \setminus \paren {\lambda F}$ is open.

So $\lambda F$ is closed.

$\blacksquare$

Proof 2

Define a mapping $c_\lambda : X \to X$ by:

$\map {c_\lambda} x = \lambda x$

for each $x \in X$.

From Dilation Mapping on Topological Vector Space is Homeomorphism, $c_\lambda$ is a homeomorphism.

From Definition 4 of a homeomorphism, $c_\lambda$ is therefore a closed mapping.

Hence $c_\lambda \sqbrk F = \lambda F$ is closed.

$\blacksquare$