Dilation of Closure of Set in Topological Vector Space is Closure of Dilation
Theorem
Let $F$ be a topological field.
Let $X$ be a topological vector space over $F$.
Let $A \subseteq X$.
Let $\lambda \in F \setminus \set {0_F}$.
Then we have:
- $\lambda A^- = \paren {\lambda A}^-$
where $A^-$ denotes the closure of $A$.
Proof 1
From Set Closure as Intersection of Closed Sets, we have:
- $\ds A^- = \bigcap \leftset {K \supseteq A: K}$ is closed in $\rightset X$
and:
- $\ds \paren {\lambda A}^- = \bigcap \leftset {K \supseteq \lambda A: K}$ is closed in $\rightset X$
For brevity write:
- $\ds \SS_1 = \leftset {K \supseteq A: K}$ is closed in $\rightset X$
and:
- $\ds \SS_2 = \leftset {K \supseteq \lambda A: K}$ is closed in $\rightset X$
so that:
- $\ds A^- = \bigcap_{K \mathop \in \SS_1} K$
and:
- $\ds \paren {\lambda A}^- = \bigcap_{K' \mathop \in \SS_2} K'$
From Dilation of Intersection of Subsets of Vector Space, it now suffices to show that:
- $\SS_2 = \set {\lambda K : K \in \SS_1}$.
Let $K \in \SS_1$.
Then $K$ is closed and $A \subseteq K$.
From Dilation of Closed Set in Topological Vector Space is Closed Set, $\lambda K$ is closed.
Since $\lambda A \subseteq \lambda K$, we have:
- $\set {\lambda K: K \in \SS_1} \subseteq \SS_2$
Now let $K' \in \SS_2$.
Then $K'$ is closed and $\lambda A \subseteq K'$.
Let $K = \lambda^{-1} K'$.
From Dilation of Closed Set in Topological Vector Space is Closed Set, $K$ is closed.
We have $A \subseteq K$, so $K \in \SS_1$.
Since we have $K' = \lambda K$ for $K \in \SS_1$, we have:
- $K' \in \set {\lambda K: K \in \SS_1}$
giving:
- $\SS_2 \subseteq \set {\lambda K: K \in \SS_1}$
So:
- $\SS_2 = \set {\lambda K: K \in \SS_1}$
by the definition of set equality.
So:
\(\ds \paren {\lambda A}^-\) | \(=\) | \(\ds \bigcap_{K' \mathop \in \SS_2} K'\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \bigcap_{K \mathop \in \SS_1} \paren {\lambda K}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lambda \bigcap_{K \mathop \in \SS_1} K\) | Dilation of Intersection of Subsets of Vector Space | |||||||||||
\(\ds \) | \(=\) | \(\ds \lambda A^-\) |
$\blacksquare$
Proof 2
Let $x \in \lambda A^-$.
Then $\lambda^{-1} x \in A^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $A$ converging to $\lambda^{-1} x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda y_\mu}_{\mu \mathop \in \Lambda}$ converges to $x$.
Since $\lambda y_\mu \in \lambda A$ for each $\mu \in \Lambda$, we have that $x \in \paren {\lambda A}^-$ from Point in Set Closure iff Limit of Net.
So $\lambda A^- \subseteq \paren {\lambda A}^-$.
Now let $x \in \paren {\lambda A}^-$.
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {y_\mu}_{\mu \mathop \in \Lambda}$ valued in $\lambda A$ converging to $x$.
From Scalar Multiple of Convergent Net in Topological Vector Space is Convergent, $\family {\lambda^{-1} y_\mu}_{\mu \mathop \in \Lambda}$ converges to $\lambda^{-1} x$.
Since $\lambda^{-1} y_\mu \in A$ for each $\mu \in \Lambda$, we have that $\lambda^{-1} x \in A^-$ from Point in Set Closure iff Limit of Net..
So $x \in \lambda A^-$.
Hence we have $\paren {\lambda A}^- \subseteq \lambda A^-$.
Hence we conclude that $\lambda A^- = \paren {\lambda A}^-$.
$\blacksquare$