# (Proof verification) “Partition” of a metric space.

Before I state the proposition, I want to make the definitions clear.

Definition. Let $$(E,d)$$ be a metric space and let $$A\subseteq E$$ be a subset. Define the interior of $$A$$ as $$int(A)=\{x \in E: \exists r>0:B(x,r)\subseteq A\}$$, where $$B(x,r)=\{y\in E:d(x,y) as usual. The exterior of $$A$$ is the set $$ext(A)$$ where $$ext(A)=int(A^c)$$. The closure of $$A$$ is the set $$cl(A)=\{x\in E:\forall r>0:B(x,r)\cap A\neq \emptyset\}$$. Finally, the boundary of $$A$$ is defined as $$bd(A)=cl(A)\cap cl(A^c)$$.

Proposition. Show that $$int(A), bd(A)$$ and $$ext(A)$$ are disjoint sets of $$E$$ and their union is the whole $$E$$.

My attempt. I start showing that the sets cover $$E$$.

\begin{align} E&=(E\setminus bd(A)) \cup bd(A) & \\ &=bd(A)^c \cup bd(A) & \\ &=(cl(A) \setminus int(A))^c \cup bd(A) & (\text{by } bd(B)=cl(B)\setminus int(B))\\ &=cl(A)^c \cup int(A)\cup bd(A) & \\ &=int(A^c) \cup int(A) \cup bd(A)& (\text{by } cl(B^c)=int(B)^c )\\ &=ext(A)\cup bd(A) \cup int(A). \end{align} Hence, the desired equality.

Now, I will show the disjointness of the sets. \begin{align} x\in int(A)& \implies (\exists r>0) B(x,r)\subseteq A \\ &\implies (\exists r>0)B(x,r)\cap A^c=\emptyset \\ &\implies x\notin cl(A^c)=A^c\cup bd(A), \end{align} since $$bd(A)=bd(A^c)$$. Then $$int(A)\cap (ext(A) \cup bd(A))=\emptyset$$ (using the fact that $$int(A^c)\subseteq A^c$$, the definition of exterior, and $$B_1\subseteq B_2, B_2=\emptyset \implies B_1=\emptyset$$). It follows that both $$(int(A)\cap ext(A))$$ and $$(int(A)\cap bd(A))$$ are empty sets.

Now

\begin{align} x\in ext(A)& \implies x\in int(A^c) \\ &\implies x\notin cl(A)=A\cup bd(A). (\text{by the above result}) \end{align}

It implies that $$(ext(A)\cap bd(A))$$ is also empty. This completes the proof.

*My actual proof is not as detailed as the one I just decribed above. I hope it helped to make the proof easier to understand.

**As @drhab mentioned, the sets do not necessarily form a partition since they can be empty.

Do you agree with this proof?

• It starts with $E=E\setminus int(A)\cup int (A)$ which is not correct. – drhab Jul 1 at 18:24
• @drhab The fact that $A\subseteq E$ and $int(A) \subseteq A$ imply $int(A)\subseteq E$. Then $E=E\setminus int(A) \cup int(A)$ holds. Am I wrong? – Danmat Jul 1 at 18:39
• Then to avoid confusion it is better to write $(E\setminus int (A))\cup int (A)$. I read it wrongly as $E\setminus (int (A)\cup int (A))$. – drhab Jul 1 at 18:57
• @drhab Ok! Assuming $E=(E\setminus int(A)) \cup int(A)$, is there anything wrong with my proof? – Danmat Jul 1 at 18:59
• Not everything is okay. You conclude that $A^c\cup bd(A^c)\cup int(A)\subseteq int(A^c)\cup bd(A^c)\cup int(A)$ on base of $int(A^c)\subseteq A^c$. That is weird. – drhab Jul 1 at 19:18

## 1 Answer

We have $$E\subseteq A\cup A^{\complement}\subseteq\mathsf{cl}\left(A\right)\cup\mathsf{cl}\left(A^{\complement}\right)\subseteq E$$ hence $$E=\mathsf{cl}\left(A\right)\cup\mathsf{cl}\left(A^{\complement}\right)$$.

In general if $$E=X\cup Y$$ then also $$E=\left(X-Y\right)\cup\left(X\cap Y\right)\cup\left(Y-X\right)$$ where $$X-Y$$, $$X\cap Y$$ and $$Y-X$$ are disjoint sets.

So we can apply this on $$X=\mathsf{cl}\left(A\right)$$ and $$Y=\mathsf{cl}\left(A^{\complement}\right)$$ and conclude $$E$$ is the union of the disjoint subsets $$\mathsf{cl}\left(A\right)-\mathsf{bd}\left(A\right)$$, $$\mathsf{bd}\left(A\right)$$ and $$\mathsf{cl}\left(A^{\complement}\right)-\mathsf{bd}\left(A\right)$$.

Now let us prove that $$\mathsf{cl}\left(A\right)-\mathsf{bd}\left(A\right)=\mathsf{int}\left(A\right)$$.

Equivalent are the statements:

• $$x\in\mathsf{cl}\left(A\right)-\mathsf{bd}\left(A\right)$$

• for every $$r>0$$ we have $$B\left(x,r\right)\cap A\neq\varnothing$$ and some $$r_{0}>0$$ exists with $$B\left(x,r_0\right)\cap A=\varnothing\text{ or }B\left(x,r_0\right)\cap A^{\complement}=\varnothing$$

• for every $$r>0$$ we have $$B\left(x,r\right)\cap A\neq\varnothing$$ and some $$r_{0}>0$$ exists with $$B\left(x,r_{0}\right)\subseteq A$$

• some $$r_{0}>0$$ exists with $$B\left(x,r_{0}\right)\subseteq A$$

• $$x\in\mathsf{int}\left(A\right)$$.

The fact that the statements under first and fifth bullet are equivalent tells us that: $$\mathsf{cl}\left(A\right)-\mathsf{bd}\left(A\right)=\mathsf{int}\left(A\right)$$

This for every set $$A$$ and applying it on $$A^{\complement}$$ we find: $$\mathsf{cl}\left(A^{\complement}\right)-\mathsf{bd}\left(A^{\complement}\right)=\mathsf{int}\left(A^{\complement}\right)$$ Since evidently $$\mathsf{bd}\left(A^{\complement}\right)=\mathsf{bd}\left(A\right)$$ and by definition $$\mathsf{int}\left(A^{\complement}\right)=\mathsf{ext}\left(A\right)$$ we can rewrite this as:$$\mathsf{cl}\left(A^{\complement}\right)-\mathsf{bd}\left(A\right)=\mathsf{ext}\left(A\right)$$

Proved is now that the sets $$\mathsf{int}\left(A\right)$$, $$\mathsf{ext}\left(A\right)$$ and $$\mathsf{bd}\left(A\right)$$ are disjoint subsets that cover $$E$$.

P.S. Formally not a partition because for that it is demanded that the sets involved are not empty. This is not necessarily the case here.

edit (alternative solution):

If $$p$$ and $$q$$ are propositions then exactly one of the following statements is true:

• $$p\wedge q$$
• $$p\wedge\neg q$$
• $$\neg p\wedge q$$
• $$\neg p\wedge \neg q$$

For a fixed $$x\in E$$ let $$p$$ be the statement: $$\exists r>0[B(x,r)\subseteq A]$$ and let $$q$$ be the statement: $$\exists r>0[B(x,r)\subseteq A^{\complement}]$$.

Then the first possibility $$p\wedge q$$ immediately falls away because it implies that $$x\in A\cap A^{\complement}$$ which is absurd.

The remaining $$3$$ statements can be translated into $$x\in\mathsf{int}(A)$$, $$x\in\mathsf{ext}(A)$$ and $$x\in\mathsf{bd}(A)$$ respectively.

So for every $$x\in E$$ it is certain that it belongs to exactly one of these sets.

• Your proof is really good. Provided one has the result $int(A)=cl(A)\setminus bd(A)$, it turns out to be very straighforward! Thanks. I still don't know why $E= E\setminus int(A) \cup int(A)$ is incorrect though. – Danmat Jul 1 at 18:48
• I have added an alternative solution. – drhab Jul 1 at 19:32