7

I apologize if this is a duplicate - I looked but didn't find one.

This question is sort of a sanity check.

Let $A$, $B$ be sets and define the boundaries $\partial A$ and $\partial B$ as usual.

Is it true that both $\partial (A \cup B) \subseteq \partial A \cup \partial B$ and $\partial (A \cap B) \subseteq \partial A \cup \partial B$?

It seems obvious, and the proofs seem really easy, but I haven't seen this fact written down anywhere.

To get started on the proofs I thought to look at a point $p$ which is in neither $\partial A$ nor $\partial B$ and then show it is not in either $\partial (A \cup B)$ or $\partial (A \cap B)$ by looking at different cases where $p$ is in the interior or exterior of $A$, $B$, taking intersections of neighborhoods, etc...

Thanks a bunch!

Amzoti
  • 56,629
bryanj
  • 4,066

2 Answers2

10

$\newcommand{\bdry}{\operatorname{bdry}}\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$You can do it with inline calculations if you use the definition that $\bdry A=\cl A\cap\cl(X\setminus A)$:

$$\begin{align*} \bdry(A\cap B)&=\cl(A\cap B)\cap\cl\Big(X\setminus(A\cap B)\Big)\\ &=\cl(A\cap B)\cap\cl\Big((X\setminus A)\cup(X\setminus B)\Big)\\ &=\cl(A\cap B)\cap\Big(\cl(X\setminus A)\cup\cl(X\setminus B)\Big)\\ &=\Big(\cl(A\cap B)\cap\cl(X\setminus A)\Big)\cup\Big(\cl(A\cap B)\cap\cl(X\setminus B)\Big)\\\\ &\subseteq\Big(\cl A\cap\cl(X\setminus A)\Big)\cup\Big(\cl B\cap\cl(X\setminus B)\Big)\\\\ &=\bdry A\cup\bdry B\;, \end{align*}$$

and

$$\begin{align*} \bdry(A\cup B)&=\cl(A\cup B)\cap\cl\Big(X\setminus(A\cup B)\Big)\\ &=\cl(A\cup B)\cap\cl\Big((X\setminus A)\cap(X\setminus B)\Big)\\ &\subseteq\cl(A\cup B)\cap\Big(\cl(X\setminus A)\cap\cl(X\setminus B)\Big)\\ &=\Big(\cl A\cup\cl B\Big)\cap\cl(X\setminus A)\cap\cl(X\setminus B)\\ &=\left(\Big(\cl A\cap\cl(X\setminus A)\Big)\cup\Big(\cl B\cap\cl(X\setminus A)\Big)\right)\cap\cl(X\setminus B)\\ &=\Big(\bdry A\cap\cl(X\setminus B)\Big)\cup\Big(\cl B\cap\cl(X\setminus A)\cap\cl(X\setminus B)\Big)\\ &=\Big(\bdry A\cap\cl(X\setminus B)\Big)\cup\Big(\bdry B\cap\cl(X\setminus A)\Big)\\\\ &\subseteq\bdry A\cup\bdry B\;. \end{align*}$$

Brian M. Scott
  • 631,399
0

$\partial (A\cap B)\subseteq \partial A \cup \partial B$

If $O_x$ is an open neighbordooh of $x\in \partial ( A\cap B)$, it has non empty intersection with $A\cap B$ and $A^C\cup B^C$.

If $O_x\cap A^C\not=\emptyset$ for all open neighborhoods $O_x$, then because $O_x\cap A\not=\emptyset$ aswell, $x\in \partial A$.

If it happens that $O_x\cap A^C=\emptyset$ for some open neighborhood $O_x$ of $x$, then $O_x\cap B^C\not=\emptyset$ as $O_x\cap(A^C\cup B^C)\not=\emptyset$.

For any other $\tilde{O}_x$, as $\tilde{O}_x\cap O_x$ is an open neighborhood of $x$ and $O_x\cap \tilde{O}_x\cap A^C=\emptyset$, then $O_x\cap \tilde{O}_x\cap B^C\not=\emptyset$ and so, in particular, $\tilde{O}_x\cap B^C\not=\emptyset$ and we can say the same of $\tilde{O}_x\cap B$, from which, $x\in \partial B$.

All in all, we have concluded $\partial (A\cap B)\subseteq \partial A\cup \partial B$.

$\partial (A\cup B)\subseteq \partial A \cup \partial B$

Notice that because the definition is symmetric with respect to complements:

$$\partial A=\{x: O_x\cap A\not=\emptyset, O_x\cap A^C, \text{ for all neighborhoods $O_x$}\}=\partial A^C$$ Hence, from $\partial(A^C\cap B^C)\subseteq \partial A^C\cup \partial B^C$, one has: $$\partial (A\cup B)=\partial (A^C\cap B^C)\subseteq \partial A^C\cup \partial B^C=\partial A\cup \partial B$$

Kadmos
  • 3,243