> [!theorem] > > Let $\cx$ be a [[Normed Vector Space|normed space]] and $E \subset X$ be any set, then > $ > \lambda \ol{E} + x = \ol{\lambda E + x} \quad \lambda E^o + x = \paren{\lambda E + x}^o > $ > for any $\lambda \ne 0$ and $x \in \cx$. > > *Proof*. First consider dilation. Let $\lambda \ne 0$, $\varepsilon > 0$, and $x \in \ol{E}$ be an [[Topological Closure|adherent point]] of $E$. Then there exists $y \in E$ such that $\norm{x - y} < \varepsilon/\abs{\lambda}$, as $\lambda y \in \lambda E$ > $ > \norm{\lambda x - \lambda y} = \abs{\lambda}\norm{x - y} < \varepsilon > $ > Therefore $\lambda x \in \ol{\lambda E}$ and $\lambda \ol{E} \subset \ol{\lambda E}$. Applying the same argument on $\ol{\lambda E}$ itself, > $ > \begin{align*} > \lambda ^{-1}\ol{\lambda E} &\subset \ol{\lambda^{-1}\lambda E} \\ > \lambda ^{-1}\ol{\lambda E} &\subset \ol{E} \\ > \ol{\lambda E} &\subset \lambda \ol{E} > \end{align*} > $ > > > Let $x \in E^o$, then there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset E$. Since > $ > B(\lambda x, \lambda \varepsilon)= \lambda B(x, \varepsilon) \subset \lambda E > $ > we have $\lambda x \in (\lambda E)^o$ and $\lambda E^o \subset (\lambda E)^o$. Applying the argument on $(\lambda E)^o$ yields > $ > \begin{align*} > \lambda^{-1}(\lambda E)^o &\subset (\lambda^{-1}\lambda E)^o \\ > \lambda^{-1}(\lambda E)^o &\subset E^o \\ > (\lambda E)^o &\subset \lambda E^o > \end{align*} > $ > > Now consider addition. Let $x \in \cx$, $\varepsilon > 0$, and $y \in \ol{E}$, then there exists $z \in E$ such that $\norm{y - z} < \varepsilon$, and $\norm{(y + x) - (z + x)} < \varepsilon$. Therefore $x + y \in \ol{E + x}$ and $\ol{E} + x \subset \ol{E + x}$. Applying the idea again yields > $ > \begin{align*} > \ol{E + x} - x &\subset \ol{E + x - x} \\ > \ol{E + x} - x &\subset \ol{E} \\ > \ol{E + x} &\subset \ol{E} + x > \end{align*} > $ > Let $y \in E^o$, then there exists $B(y, \varepsilon) \subset E^o$ and $B(y + x, \varepsilon) \subset E + x$. Therefore $E^o + x \subset (E + x)^o$, and > $ > \begin{align*} > (E + x)^o - x &\subset (E + x - x)^o \\ > (E + x)^o - x&\subset E^o \\ > (E + x)^o &\subset E^o + x > \end{align*} > $