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 $\lambda \in K$.

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

Proof
First suppose that $\lambda = 0$.

Then we have $\lambda F = \set { {\mathbf 0}_X}$.

From Finite Topological Space is Compact, $\lambda F$ is then a compact subspace of $X$.

Since $X$ is a topological vector space, it is Hausdorff, and so $\lambda F$ is closed from Compact Subspace of Hausdorff Space is Closed.

Now suppose that $\lambda \ne 0$.

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

We show that:


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

It is immediate from the definition of a dilation that if $x \in F$ we have $\lambda x \in \lambda F$.

Conversely, if $\lambda x \in \lambda F$, we have $\lambda x = \lambda y$ for some $y \in F$.

That is, $\lambda \paren {x - y} = 0$.

Since $\lambda \ne 0$, it follows that $x = y$, and so $x \in F$.

So we have $x \in F$ $\lambda x \in \lambda F$.

So for $x \in X$ we have $x \not \in F$ $\lambda x \not \in \lambda F$.

That is, $x \in X \setminus F$ $\lambda x \in X \setminus \paren {\lambda F}$.

It follows that $\lambda x \in X \setminus F$ $\lambda x \in X \setminus \paren {\lambda F}$, giving the equality:


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

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.