Prove that $cl(M)=int(M)\cup Bd(M)$ in an algebraic way I need help with my following problem. We defined closure of a set as

$cl(M)= M \cup Bd(M)$ with $M\subset \mathbb{R^n}$

But I have seen it also this way

$cl(M)= int(M) \cup Bd(M)$. This is what I want to prove!

Firt of all let me make some things clear: We defined cl(M) as the closure of M with $cl(M)= M \cup Bd(M)$. $Bd(M)$ is the set of all boarder points of M. A point is a boarder point of M, if for each $\epsilon$ $>0$ there is a open Ball with radius $\epsilon$ around this point, that contains one or more elements of M and $M^c$.
We defined the interior of M int(M)= $M\setminus Bd(M)$
So now my problem: I tried to prove

$cl(M)= int(M) \cup Bd(M)$

in an algebraic way, but it doesn't work for me and unfortunately I don't find any post, that helps me solve my problem the way I like it. Actually there is one post, that suggests to use $Bd(M)=cl(M)\setminus int(M)$, but it does not lead to anything: Let $A$ be a subset of a topological space. Prove that $Cl(A) = Int(A) \cup Bd(A)$
Lets take $$cl(M)=int(M)\cup Bd(M) \implies cl(M)=int(M) \cup (cl(M)\setminus int(M)) $$
Someone in the post, I linked above, then claimed, that this leads us to the fact that  $cl(M)=cl(M)$, and thats why it is true to say $cl(M)=int(M) \cup Bd(M)$. But how did he get cl(M) on the right side. If I continue I get
$$cl(M)=(int(M)\cup cl(M)) \cap (int(M) \cup int(M)^c) \implies cl(M)=(int(M)\cup cl(M))\cap \mathbb{R^n}$$ $$\implies cl(M)= int(M)\cup cl(M)$$
and thus, I am again at the beginning and I didn't prove anything.
Did I do a mistake? Or should I approach differentely? Is there anyone who could help me out? I would be very grateful.
 A: One inclusion,  $\overline M\supset {M}^{\circ}\cup \partial M$, is easy.
For the other,  let $x\in\overline M$.  Then if $x\not\in M^\circ$, then every open set $U$ containing $x$ intersects $M^c$.  Thus $x\in\overline{(M^c)}$ .  Thus $x\in\partial M$.
A: Denote the closure $cl(M)=\bar{M}$, the interior $int(M)=\overset{o}{M}$, and the boundary $Bd(M)=\partial M$.
If we start from
$$ \partial M=\bar{M} \cap(\overset{o}{M})^c.$$
Then by adding $\overset{o}{M}$ to both sides we get
\begin{align*}
\partial M \cup \overset{o}{M}&=\left(\bar{M} \cap(\overset{o}{M})^c\right)\cup \overset{o}{M}\\
&=\left(\bar{M}\cup M\right)\cap\left((\overset{o}{M})^c\cup \overset{o}{M}\right)\\ 
\end{align*}
We have $\overset{o}{M}\subseteq M\subseteq\bar{M}$. Using De Moivre's property, we get
\begin{align*}
\partial M \cup \overset{o}{M}&=\bar{M}\cap E\\ 
&=\bar{M}
\end{align*}
where $E$ represents the total space ($\mathbb R^n$ in your case).
A: To show $cl(M)= int(M) \cup Bd(M)$
Notation : $cl(M) =\overline{M}$ , $int(M)=\overset{o}{M}$ and $Bd(M) =\partial{M}$
Hints :$\overset{o}{M}\subset M\subset \overline{M}$ , $\partial{M}=\overline{M} \setminus \overset{o}{M}$ and $A \cup (B\setminus C) =(A \cup B)\setminus (C\setminus A)$

$\begin{align}\overset{o}{M}\cup \partial{M}&=\overset{o}{M} \cup( \overline{M} \setminus \overset{o}{M})\\&=  (\overset{o}{M}\cup \overline{M})\setminus (\overset{o}{M}\setminus \overset{o}{M})\\&=\overline{M}\setminus{\emptyset} \\&=\overline{M}\end{align} $
A: Given a metric space $(X,d_{X})$ and a subset $M\subseteq X$, you can prove that $p\in X$ is an adherent point of $M$ if, and only if, it is an interior point of $M$ or a boundary point of $M$ as you have suggested. In order to do so, we shall propose the following equivalent sentences:

*

*$p\in X$ is an adherent point of $M$

*$p\in X$ is an interior or boundary point of $M$

*There exists a sequence of points $p_{n}\in M$ which converges to $p$
Let us prove the first that the first statement implies the second.
Suppose that $p\in X$ is an adherent point of $M$ as well as an exterior point. Well, this leads to a contradiction, because (by definition of exterior point) there is an open ball $B_{\varepsilon}(p)\subseteq M^{c}$, which implies that $B_{\varepsilon}(p)\cap M = \varnothing$.
We shall now prove that the second statement implies the third.
If $p\in\text{int}(M)$, then $p\in M$ by the definition involved. Then it suffices to take $p_{n} := p$.
If $p\in\partial M$, then it is neither an interior point nor an exterior point. In particular, the following relation holds: $B_{\varepsilon}(p)\not\subset M^{c}$ for every $\varepsilon > 0$. This means that, for every $\varepsilon > 0$, one has that $B_{\varepsilon}(p)\cap M\neq\varnothing$. In particular, if we choose $\varepsilon = 1/n$, there corresponds some $p_{n}\in B_{1/n}(p)\cap M \subseteq B_{1/n}(p)$ which means that $d_{X}(p_{n},p) < 1/n$. Taking the limit from both sides, one gets the desired claim.
Finally, we shall prove the third statement implies the first.
Let us remind the definition of convergence in metric spaces. We say the sequence of points $(p_{n})_{n\in\mathbb{N}}$ in $(X,d_{X})$ converges to $p\in X$ iff
\begin{align*}
(\forall\varepsilon\in\mathbb{R}_{>0})(\exists n_{\varepsilon}\in\mathbb{N})(\forall n\in\mathbb{N})(n\geq n_{\varepsilon} \Rightarrow d_{X}(p_{n},p) < \varepsilon)
\end{align*}
Consequently, based on the assumption of convergence, no matter which $\varepsilon\in\mathbb{R}_{>0}$ one chooses, there is always some $n_{\varepsilon}\in\mathbb{N}$ such that $d_{X}(p_{n},p) < \varepsilon$ whenever $n\geq n_{\varepsilon}$. We can deduce from the last statement that: no matter which $\varepsilon\in\mathbb{R}_{>0}$ one chooses, there is at least $p_{n_{\varepsilon}}\in B_{\varepsilon}(p)$, and we are done.
Final comments
If you are interested in the topological proof of such result, I think it suffices to substitute open balls by open neighborhoods.
