Exercise 2.9

Question

Exercise 9: (a) Prove that $E^\circ$ is always open. (b) $E$ is open iff $E = E^\circ$. (c) If $G \subset E$ is open, then $G \subset E^\circ$. (d) Prove that $ (E^\circ)^c = \overline{E^c}$. (e) Do $E$ and $\bar E$ always have the same interiors? (f) Do $E$ and $E^\circ$ always have the same closures?

ghostofgarborg · Accepted Answer

By definition, if $x \in E^\circ$, then $x$ in an open subset $U \subset E$. Let $U \subset E$ be open. Then any point of $U$ has a neighborhood inside $E$, so that 
    $ U \subset E^\circ $. This establishes that 
    $$ E^\circ = \bigcup_{\substack{U \subset E \ U \text{ open}}} U $$
    (a) $E^\circ$ is a union of open sets, and is therefore open. (Note: This is also true if the union is empty.)

(b) If $E = E^\circ$, then we know from (a) that $E$ is open. 
    For the converse, begin by noting that $ E^\circ \subseteq E$, since it is a union of subsets of $E$. 
    If $E$ is open, then $E$ is in the union, so that 
    $E \subseteq E^\circ$ as well, from which equality follows.

(c) This is clear, as $G$ is included in the union.

(d) By theorem 2.27 (a) and (e) 
    $$ (E^\circ)^c = \bigcap_{\substack{U \subset E \ U \text{ open}}} U^c 
        = \bigcap_{\substack{V \supset E^c \ V \text{ closed}}} V 
        = \overline{E^c} $$

(e) No. Let $E = \mathbb Q$ in $\mathbb R$. Then $ E^\circ = \emptyset$, but $(\bar E)^\circ = \mathbb R^\circ = \mathbb R$.

(f) No. Let $E = \mathbb Q$ in $\mathbb R$. Then $ \bar E = \mathbb R$, but $ \overline{ E^\circ } = \bar \emptyset = \emptyset$.

Exercise 2.9

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Exercise 2.9

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