5

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
  • 56,093
bryanj
  • 3,938

1 Answers1

6

$\newcommand{\bdry}{\operatorname{bdry}}\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$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
  • 616,228