I am trying to prove that every neighborhood of a boundary bound contains a point in interior and $X \setminus A$ where $A$ is the set in consideration. I am given the following definitions
(1) $(X,\mathcal{T})$ is a topological space if $X,\emptyset \in \mathcal{T}$, any arbitrary union of open sets in $\mathcal{T}$ is an open set, and any finite intersection of an open set in $\mathcal{T}$ is an open set.
(2) A neighborhood of a subset of $X$ is an open set containing the subset.
(3) A set $A \subseteq X$ is closed if $X \setminus A$ is open
(4) $\bar A$ = $\cap\{ B \subseteq X : A \subseteq B \text{ and } B \text{ is closed in X} \}$
(5) Int($A$) = $\cup \{ B \subseteq A : B \text{ is open in } X \}$
(6) Ext($A$) = $X \setminus \bar A$
(7) $\partial A$ = $X \setminus ( \text{Int(}A)\text{)}\cup \text{Ext(}A \text{)})$
From this I can derive the closed definition of a topology and that $\partial A$ is closed. I am having trouble seeing how the previous facts and definitions give me that every neighborhood of a point in the boundary contains a point in $A$ and $X \setminus A$ where $X$ is the space and $A$ is some subset of the space. What are some derivations that would help me figure this out?
What I think it comes down to is showing that given an open neighborhood $\mathcal{O}$ of some point $x \in \partial A$, $\mathcal{O} \cap Int(A)\ne \emptyset$ and $\mathcal{O} \cap (X \setminus A) \ne \emptyset$
Best Answer
Observe that if $x$ is in the boundary of $A$, then $x$ is not in the interior or the exterior of $A$. Since $x\notin\operatorname{int}A$, $x$ has no open nbhd contained entirely in $A$. But that just means that if $U$ is an open nbhd of $x$, then $U\nsubseteq A$, and hence $U\cap(X\setminus A)\ne\varnothing$. A similar argument shows that since $x$ is not in the exterior of $A$, every open nbhd of $x$ must meet $A$ in a non-empty set. The converse is equally easy to show: if every open nbhd of $x$ meets both $A$ and $X\setminus A$, then $x$ is in the boundary of $A$. To complete the argument, prove the following proposition: