I'm trying to prove the following statement. I can prove it with words, but I'm trying to see how the logic checks out.
Let $(\Omega ,d)$ be a metric space. Assume the following definitions of the set $E$, the closure$\bar{E}$, the derived set $E'$ (the set of all accumulation points of E), and the boundary $\partial E$:
$$E= \left \{ x\in\Omega :x \in E \right \}$$
$$\bar{E}=E\, \cup\, E' $$
$$E'=\left \{ x\in\Omega :(B_r(x)\setminus \left \{ x \right \})\cap E\neq \varnothing \;\; \forall r \in \mathbb{R}_+ \right \}$$
$$\partial E=\left \{ x\in\Omega :(B_r(x)\cap E\neq \varnothing\:) \wedge (B_r(x)\cap E^c\neq \varnothing\:) \;\; \forall r \in \mathbb{R}_+ \right \}$$
where $B_r(x)$ is the open ball of radius $r$ centered at $x \in \Omega $.
I want to prove that $\bar{E}=E \, \cup \, \partial E$ by seeing if I can use logic to derive the set $E \cup \partial E$ starting from the definition $\bar{E}=E\, \cup\, E' $, but I'm having a bit of trouble with the logic.
Using the idempotent property of intersection, I wrote
$$\bar{E}=E\, \cup\, E'= E\, \cup\,(E' \cap E')=E\, \cup\, \left \{ x\in\Omega :((B_r(x)\setminus \left \{ x \right \})\cap E\neq \varnothing) \wedge ((B_r(x)\setminus \left \{ x \right \})\cap E\neq \varnothing)) \;\; \forall r \in \mathbb{R}_+ \right \} $$
But I want
$$E \cup \partial E = E \cup \left \{ x\in\Omega :(B_r(x)\cap E\neq \varnothing\:) \wedge (B_r(x)\cap E^c\neq \varnothing\:) \;\; \forall r \in \mathbb{R}_+ \right \}$$
and I'm not exactly sure where to go with the logic. Part of me thinks that the right direction is to write
$$(B_r(x)\setminus \left \{ x \right \})\cap E=B_r(x)\cap \left \{ x \right \}^c\cap E=B_r(x)\cap (\Omega \setminus \left \{ x \right \})\cap E=B_r(x)\cap ((E\cup E^c) \cap \left \{ x \right \}^c)\cap E$$
and go from there since I know that one of the sets $E \cap \left \{ x \right \}^c $ and $E^c \cap \left \{ x \right \}^c$ has to be empty, but I don't think that really gets me anywhere.