Prove a claim on the boundary of A ($\partial A$)

Question

Let $A\subseteq \mathbb R^n$ (A is a Set), and let $x \in \partial A$ (in other words; x is a limit point)

Note: $\partial A$ refers to the boundary of $A$. link: https://en.wikipedia.org/wiki/Boundary_(topology)

I want to prove the following:

p is a boundary point of a set if and only if every neighborhood of p contains at least one point in the set and at least one point not in the set.

I am trying to prove this by contradiction; So I supposed:

1) There is a a boundary point x of the set A such that there is a neighborhood of x that contains only points from the set.

Or

2) There is a a boundary point x of the set A such that there is a neighborhood of x that contains only points not from the set.

I proved that 1 is wrong, but how could I prove that the second is wrong too?

Answer 1

You need to argue as follows: Let $x\in\partial A$, and consider a neighborhood $N$ of $x$ such that $N\cap A=\emptyset$.

Then $x\not\in closure(A)$: simply because $N^c$ is closed, $A\subset N^c$, and $x\not\in N^c$. A simple argument then allows you to conclude that $x\not\in\partial A$. [Hint: if $x$ is not in the closure of $A$, then it must belong to the exterior of $A$ and $\partial A = (int(A)\cup ext(A))^c$]

Answer 2

Just for the sake of curiosity, I provide here an alternative way to approach it.

Let $(X,d_{X})$ be a metric space, $E\subseteq X$ and $x\in X$. We say that $x$ is a boundary point of $E$ if it is neither an interior point nor an exterior point of $E$.

We say that $x\in\text{int}(E)$ iff there exists a positive real number $r > 0$ s.t. $B(x,r)\subseteq E$.

We say that $x\in\text{ext}(E)$ iff there exists a positive real number $R > 0$ s.t. $B(x,R)\cap E = \varnothing$.

Thus $x\in X$ is in the boundary of $E$ iff $B(x,r)\not\subseteq E$ and $B(x,r)\cap E\neq\varnothing$ for every $r > 0$.

The last statements prove what you are looking for.