If $x$ is a limit point of $A$, then $x$ is either an interior or a boundary point of $A$

Question

"Prove that if $x$ is a limit point of $A$, then $x$ is either an interior or a boundary point of $A$"

There are two options. Either $x\in A$ or $x\in A^C$. Since $x$ is a limit point, there exists neighborhoods $d(x,y)<\epsilon/2$ and $d(x,z)<\epsilon/2$. It follows that $$d(y,z)\leq d(y,x)+d(x,z)<\epsilon$$

This is where I'm starting to doubt my logic. This shows that there is a neighborhood of $x$ which is contained in the neighborhood of an arbitrary point $y\in A$, because $N_r(y)$ is just the interval $$0\leq d(y,z)<\epsilon $$

and clearly some $d(x,y)$ is contained within. Can we claim from these facts, that $x$ is an interior point of $A$?

Considering $x \in A^C$. Now every neighborhood of $x$ contains a point $t\in A$, such that $t\neq x$. A neighborhood of a point contains the point itself, and it's clear that every $N_\epsilon (x)$ now contains $x\in A^C$ and some $t\in A$. Therefore $x$ is a boundary point of $A$.

Answer 1

There are two mutually exclusive logical options for a limit point $x$ of $A$:

1. $\exists r>0: B(x,r) \subseteq A$.

2. $\forall r>0: B(x,r) \cap A^\complement \neq \emptyset$.

In $1$ $x$ is an interior point of $A$ and in $2$ $x$ is a boundary point of $A$ (as every ball intersects $A$ (from the limit point property) and $A^\complement$.

Answer 2

It's easier to prove that $x$ is not an outer point of $A$. That's because for every neighbourhood of the point you will always find some points of the sequence inside that neighbourhood. As those points belong to $A$, $x$ cannot be an outer point.