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 $B(x,r)=\{y\in E:d(x,y)<r\}$ 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.


\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?

  • $\begingroup$ It starts with $E=E\setminus int(A)\cup int (A)$ which is not correct. $\endgroup$ – drhab Jul 1 at 18:24
  • $\begingroup$ @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? $\endgroup$ – Danmat Jul 1 at 18:39
  • $\begingroup$ Then to avoid confusion it is better to write $(E\setminus int (A))\cup int (A) $. I read it wrongly as $E\setminus (int (A)\cup int (A))$. $\endgroup$ – drhab Jul 1 at 18:57
  • $\begingroup$ @drhab Ok! Assuming $E=(E\setminus int(A)) \cup int(A)$, is there anything wrong with my proof? $\endgroup$ – Danmat Jul 1 at 18:59
  • $\begingroup$ 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. $\endgroup$ – drhab Jul 1 at 19:18

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.

  • $\begingroup$ 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. $\endgroup$ – Danmat Jul 1 at 18:48
  • $\begingroup$ I have added an alternative solution. $\endgroup$ – drhab Jul 1 at 19:32

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