A closed set contains all its boundary points. Is there a first principle method for proving that a closed set contains all its boundary points, based only on the definitions of the closed set and boundary points, without using the former's complementarity with the corresponding open set?
PROBLEM: Consider a closed set $A \subseteq \mathbb{R^{n}}$. Show that if $x^{0}$ is a boundary point of A, then $x^{0}$ belongs to A.
I have come up with the following proof:
PROOF: It is given that $x^{0}$ is a boundary point of $A$, i.e., every open-ball $V_{x^{0}}$ around $x^{0}$ contains points both inside and outside $A$. Of the points inside $A$ lying in $V_{x^{0}}$ , a sequence $\{x^{k}\}$ can be constructed that converges to $x^{0}$ as $k \to \infty$. Since $A$ is a closed set, then $x^{k} \to x^{0} \implies  x^{0} \in A$. Hence, proved.
Is this proof correct? Can we assert that there would be a sequence in $A\cap V_{x^{0}}$ that would converge to $x^{0}$ in the first place?
 A: Your proof is correct in the context of metric spaces.
We can also prove this in the more general context of topological spaces by replacing open balls with neighborhoods. Let the closed set be $S$ in the topological space $X$. Let $x\in\partial S$. Suppose to the contrary, we have found $x$ such that $x\notin S$.
Since $x\in\partial S$, every neighborhood of $x$ has an element of $S$. Since we assume $x\notin S$, these elements are distinct from $x$ itself. Therefore, $x$ is a limit point of $S$. But closed sets contain their limit points, so $x\in S$. Contradiction.
So one cannot find any points in $\partial S\setminus S$.
A: You can use several characterisations of closed sets to show this fact:

*

*$A$ is closed iff it contains all its limit points (or $A' \subseteq A$), where a limit point $x$ of $A$ is defined by the fact that any neighbourhood of $x$ intersects $A\setminus \{x\}$. If you know this, it's trivial because a boundary point of $A$ is either already in $A$ or a limit point of $A$.


*$A$ is closed iff $x \in A$ iff $x$ is an accumulation point of $A$: i.e. every neighbourhood of $x$ intersects $A$ (closely related notion to limit point; the set of accumulation points of $A$ is also denoted by $\overline{A}$ and called the closure of $A$). In that case, it's also trivial as any boundary point of $A$ is a fortiori an accumulation point (and thus in $\overline{A}$ which equals $A$, for $A$ closed).


*$A$ is closed iff every convergent sequence from $A$ has its limit in $A$. This characterisation only holds in so-called sequential spaces $X$, but these include all metric spaces, so you're safe there. This is essence your argument AFAICS: if $x \in \partial A$ you construct $a_n \in A$ so that $a_n \to x$ (in fact this can be done in any metric space with any accumulation or limit point of $A$!) and then apply the fact. It's valid, but a bit indirect, compared to the previous two approaches.
All of the above is equivalent to defining closed sets as the complements of open sets and the closure as the smallest closed set containing $A$ etc. This is classical and proved many times over on this web site alone...
