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

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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 $\lambda A^- = \paren {\lambda A}^-$.

$\blacksquare$