4. Indexed families

Part 1.: Problems, solutions.

Indexed families

An indexed family is a family (set of sets), but such that its elements (sets) are indexed (e.g. by natural numbers). Formally an indexed family is a function from the set of indices into the set of its elements.

E.g. A_n=\{[0,n]\colon n\in\mathbb{N}\} is a family of intervals of the real line, indexed by natural numbers.

Unions and intersections of indexed families

We will be especially interested in unions and intersections of indexed families. We calculate those unions and intersection similarly to the case of simple families of sets, but this time we can include information which indices we would like to use.

For the family in the above example: \bigcup_{n\in \mathbb{N}}A_n=\bigcup_{n=0}^{\infty}A_n=[0,+\infty), because it is clear that no negative number is in this union. On the other hand, if r\in[0,+\infty) is a real number, then let n=\lceil r\rceil. Then r\in A_n=[0,n], and so r\in\bigcup_{n=0}^\infty A_n.

On the other hand \bigcap_{n\in\mathbb{N}}A_n=\{0\}, because for all n, 0\in A_n=[0,n] and \bigcap_{n\in\mathbb{N}}A_n\subseteq A_0=\{0\}.

Meanwhile, e.g. \bigcap_{n=5}^\infty A_n=[0,5], and \bigcup_{n=0}^{2015}=[0,2015].

Double indexed families

Obviously the set of indices can be a product of two sets, so sets in the family can be indexed by two indices. E.g.: A_{n,m}=[-n,m], n,m\in\mathbb{N}.

Therefore we can imagine such a family as an infinite array of sets:

    \[\begin{array}{c|cccc} &n=0&n=1&n=2&\cdots\\ \hline m=0&\{0\}&[-1,0]&[-2,0]&\cdots\\m=1&[0,1]&[-1,1]&[-2,1]&\cdots\\m=2&[0,2]&[-1,2]&[-2,2]&\cdots\\\cdots&\cdots&\cdots&\cdots&\end{array}\]

Double unions and intersections

Given a double indexed family, the expressions of the following type make sense: e.g. \bigcup_{n=0}^\infty\bigcap_{m=0}^\infty A_{n,m}, czy \bigcap_{m=0}^\infty\bigcup_{n\geq m}^\infty A_{n,m} and other similar. Let us calculate those two expressions.

To calculate the first expression, denote B_n=\bigcap_{m=0}^\infty A_{n,m} — it is the intersection of all the sets in the n-th column of this infinite array. It is clear that B_n=[-n,0], because that for any m, [-n,0]\subseteq A_{n,m} and A_{n,0}=[-n,0], so B_n\subseteq[-n,0]. Therefore, \bigcup_{n=0}^\infty\bigcap_{m=0}^\infty A_{n,m}=\bigcup_{n=0}^\infty B_n = \bigcup_{n=0}^\infty [-n,0]=(-\infty,0], because elements of all the sets in this union are less than or equal to zero and for all x\leq 0, x\in B_n for n=\lceil -x\rceil.

To calculate the second expression, denote C_m=\bigcup_{n\geq m}^\infty A_{n,m}, so it is the union of all sets in the m-th row starting with the m-th element. So C_m=(-\infty,m], because for any n\geq m, A_{n,m}\subseteq(-\infty,m] and if 0\leq x\leq m, then x\in A_{m,m}\subseteq C_m. On the other hand if x<0, then x\in A_{n,m}, for n=max(m,\lceil-x\rceil). Therefore, \bigcap_{m=0}^\infty\bigcup_{n\geq m}^\infty A_{n,m}= \bigcap_{m=0}^\infty C_m= \bigcap_{m=0}^\infty (-\infty, m]=(-\infty,0]. That is because for any m, (-\infty,0]\subseteq C_m and C_0=(-\infty,0], and therefore \bigcap_{m=0}^\infty C_m\subseteq (-\infty,0].