Dilation of Interior of Set in Topological Vector Space is Interior of Dilation

Theorem

Let $K$ be a topological field.

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

Let $A \subseteq X$.

Let $\lambda \in K \setminus \set 0$.

Then we have:

$\lambda A^\circ = \paren {\lambda A}^\circ$

where $A^\circ$ denotes the intrerior of $A$.

Proof

From the definition of intrerior we have:

$\ds A^\circ = \bigcup \leftset {U \subseteq A: U}$ is open in $\rightset X$

and:

$\ds \paren {\lambda A}^\circ = \bigcup \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$

For brevity write:

$\ds \SS_1 = \leftset {U \subseteq A: U}$ is open in $\rightset X$

and:

$\ds \SS_2 = \leftset {U \subseteq \lambda A: U}$ is open in $\rightset X$

so that:

$\ds A^\circ = \bigcup_{U \mathop \in \SS_1} U$

and:

$\ds \paren {\lambda A}^\circ = \bigcup_{U \mathop \in \SS_2} U$

From Dilation of Union of Subsets of Vector Space, it now suffices to show that:

$\SS_2 = \set {\lambda U : U \in \SS_1}$

Let $U \in \SS_1$.

Then $U$ is open and $U \subseteq A$.

From Dilation of Open Set in Topological Vector Space is Open, $\lambda U$ is open.

Since $\lambda U \subseteq \lambda A$, we have:

$\set {\lambda U: U \in \SS_1} \subseteq \SS_2$

Now let $U' \in \SS_2$.

Then $U'$ is open and $U' \subseteq \lambda A$.

Let $U = \lambda^{-1} U'$.

From Dilation of Open Set in Topological Vector Space is Open, $U$ is open.

We have $U \subseteq A$, so $U \in \SS_1$.

Since we have $U' = \lambda U$ for $U \in \SS_1$, we have:

$U' \in \set {\lambda U: U \in \SS_1}$

giving:

$\SS_2 \subseteq \set {\lambda U : U \in \SS_1}$

So:

$\SS_2 = \set {\lambda U : U \in \SS_1}$

by the definition of set equality.

So:

\(\ds \paren {\lambda A}^\circ\)

\(=\)

\(\ds \bigcup_{U' \mathop \in \SS_2} U'\)

\(\ds \)

\(=\)

\(\ds \bigcup_{U \mathop \in \SS_1} \paren {\lambda U}\)

\(\ds \)

\(=\)

\(\ds \lambda \bigcup_{U \mathop \in \SS_1} U\)

Dilation of Union of Subsets of Vector Space

\(\ds \)

\(=\)

\(\ds \lambda A^\circ\)

$\blacksquare$