Dilation of Closure of Set in Topological Vector Space is Closure of Dilation/Proof 2

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