Exercise 3.2.14

Question

A dual notion to the closure of a set is the \emph{interior} of a set. The interior of $E$ is denoted $E^{\circ}$ and is defined as
  $$
  E^{\circ}=\left\{x \in E: 	ext { there exists } V_{\epsilon}(x) \subseteq Eight\}
  $$
  Results about closures and interiors possess a useful symmetry.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that $E$ is closed if and only if $\overline{E}=E .$ Show that $E$ is open if and only if $E^{\circ}=E$.
  \item Show that $\overline{E}^{c}=\left(E^{c}ight)^{\circ}$, and similarly that $\left(E^{\circ}ight)^{c}=\overline{E^{c}}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item
    \begin{enumerate}[label=(oman*)] 
    \item If $E = \overline{E} = E$.
    \item If $E^\circ = E$ then every $x \in E$ has $V_\epsilon(x) \subseteq E$ therefore $E$ is open. If $E$ is open then every $x \in E$ has $V_\epsilon(x) \subseteq E$ therefore $E = E^\circ$.
     \end{enumerate}
  \item $x \in \overline{E}^c$ iff $x 
otin E$ and $x$ is not a limit point of $E$, $x \in (E^c)^\circ$ iff $x 
otin E$ and there exists $V_\epsilon(x) \subseteq E^c$. Notice ``x is not a limit point of $E$'' is equivalent to ``there exists $V_\epsilon(x) \subseteq E^c$'' therefore the sets are the same.

To show $(E^\circ)^c = \overline{E^c}$ let $D = E^c$ yielding $((D^c)^\circ)^c = \overline{D}$ taking the complement of both sides yields $(D^c)^\circ = \overline{D}^c$ which we showed earlier.
   \end{enumerate}

Exercise 3.2.14

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Exercise 3.2.14

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