Families of Sets and `Generalized' Set Operations

Finite Set Operations

The operations \cup and \cap apply to two sets; we can however use them to combine more than one set. If we have a finite collection of sets A_1,A_2,\ldots,A_N, we write

(1)   \begin{equation*}\begin{aligned}\bigcup_{i=1}^N A_i&=A_1\cup A_2\cup \cdots\cup A_N\\\bigcap_{i=1}^N A_i&=A_1\cap A_2\cap \cdots\cap A_N\end{aligned}\end{equation*}

Activity. Give a better definition of \displaystyle\bigcup_{i=1}^NA_i and \displaystyle\bigcap_{i=1}^NA_i . (Hint. What did we say about definitions involving \cdots?)

De Morgan's Laws. For any finite collection of sets A_1,\ldots, A_n,

    \begin{align*}\left(\bigcup_{i=1}^N A_i\right)^c=&\bigcap_{i=1}^NA_i^c\\\left(\bigcap_{i=1}^N A_i\right)^c=&\bigcup_{i=1}^NA_i^c \end{align*}

Exercise. Prove this version of de Morgan's Laws.

Transfinite Operations

There is no reason to restrict our attention to {\em finite} collections of sets. For example, we might consider

the collection of intervals of the form [n,n+1], where n is allowed to be an natural number,

or perhaps

the set of all disks centered at the origin

each of which consists of an infinite number of sets. We usually use set-builder notation to write something like

(2)   \begin{equation*}\left\{[n,n+1]\middle\vert n\in\mathbb{N}\right\}\end{equation*}


(3)   \begin{equation*}\left\{D_r\middle\vert r\in(0,\infty) \right\}\end{equation*}

to emphasize that what we are dealing with is a(n infinite) set of sets. (Here we have written D_r=\left\{(x,y)\middle\vert x^2+y^2<r^2\right\} for the disk of radius r centered at the origin.) We call the set to the right of the pipe the index set: in these collections  the index sets are \mathbb{N} and the interval (0,\infty), respectively.

We can define unions and intersections of such infinite collections of sets by recalling that \exists acts like a souped-up version of \vee and \forall acts like a souped-up version of \wedge.

Given a collection of sets \mathcal{A}=\left\{A_\lambda\middle\vert \lambda\in\Lambda\right\}, we set


    \begin{align*}\bigcup\mathcal{A}&=\bigcup_{\lambda\in\Lambda}A_\lambda=\left\{x\middle\vert \exists\lambda\in\Lambda:x\in A_\lambda\right\}\\\bigcap\mathcal{A}&=\bigcap_{\lambda\in\Lambda}A_\lambda=\left\{x\middle\vert \forall\lambda\in\Lambda,x\in A_\lambda\right\}\end{align*}

E.g. \displaystyle\bigcup_{n\in\mathbb{N}} [n,n+1]=[1,\infty)

Proof. Let x\in[1,\infty). Then there is some n\in\mathbb{N} with n\leq x\leq n+1, so x\in[n,n+1]. Thus \displaystyle x\in\bigcup_{n\in\mathbb{N}} [n,n+1].

Now let \displaystyle x\in\bigcup_{n\in\mathbb{N}} [n,n+1]. Then there is some n\in\mathbb{N} with x\in[n,n+1]. But [n,n+1]\subseteq [1,\infty), so x\in[1,\infty). \Box

E.g. \displaystyle\bigcap_{r\in(0,\infty)} D_r=\{(0,0)\}

Proof. Certainly the origin (0,0) is in each disk D_r. We will show it is the only such point.

Consider any point (x,y) in the plane. Let r_1=\sqrt{x^2+y^2} be the distance between (x,y) and the origin. Then if we select r=\frac{1}{2}r_1, we see that (x,y)\notin D_r. Thus (x,y) cannot be in every D_r.

de Morgan's Laws. For any collection \{A_\lambda|\lambda\in\Lambda\}, we have

    \begin{align*}\left(\bigcup_{\lambda\in\Lambda} A_\lambda\right)^c=&\bigcap_{\lambda\in\Lambda}A_\lambda^c\\ \left(\bigcap_{\lambda\in\Lambda} A_\lambda\right)^c=&\bigcup_{\lambda\in\Lambda}A_\lambda^c \end{align*}

Exercise. Prove this version of de Morgan's Laws.

Exercise. What is each of the following sets?

  1. \displaystyle\bigcap_{n\in\mathbb{N}} [n,n+1]
  2. \displaystyle\bigcup_{r\in(0,\infty)} D_r
  3. \displaystyle\bigcup_{n\in\mathbb{N}}\left[\frac{1}{n},1\right]
  4. \displaystyle\bigcap_{n\in\mathbb{N}}\left[\frac{1}{n},1\right]

Exercise Express each set below as either a generalized union or a generalized intersection, and also express it in set-builder notation.

  1. \displaystyle B\setminus\left(\bigcap_{\lambda\in\Lambda} A_\lambda\right)
  2. \displaystyle B\setminus\left(\bigcup_{\lambda\in\Lambda} A_\lambda\right)
  3. \displaystyle \left(\bigcap_{\lambda\in\Lambda} A_\lambda\right)\setminus B
  4. \displaystyle \left(\bigcup_{\lambda\in\Lambda} A_\lambda\right)\setminus B

Published in sets.