Section 5.5 Indexed Families of Sets
Beginning Activity Beginning Activity 1: The Union and Intersection of a Family of Sets
In Section 5.3, we discussed various properties of set operations. We will now focus on the associative properties for set union and set intersection. Notice that the definition of “set union” tells us how to form the union of two sets. It is the associative law that allows us to discuss the union of three sets. Using the associate law, if \(A\text{,}\) \(B\text{,}\) and \(C\) are subsets of some universal set, then we can define \(A \cup B \cup C\) to be \(\left( A \cup B \right) \cup C\) or \(A \cup \left( B \cup C \right)\text{.}\) That is,
For this activity, the universal set is \(\N\) and we will use the following four sets:
1.
Use the roster method to specify the sets \(A \cup B \cup C\text{,}\) \(B \cup C \cup D\text{,}\) \(A \cap B \cap C\text{,}\) and \(B \cap C \cap D\text{.}\)
2.
Use the roster method to specify each of the following sets. In each case, be sure to follow the order specified by the parentheses.
(a)
\(\left( A \cup B \cup C \right) \cup D\)
(b)
\(A \cup \left( B \cup C \cup D \right)\)
(c)
\(A \cup \left( B \cup C \right) \cup D\)
(d)
\(\left( A \cup B \right) \cup \left( C \cup D \right)\)
(e)
\(\left( A \cap B \cap C \right) \cap D\)
(f)
\(A \cap \left( B \cap C \cap D \right)\)
(g)
\(A \cap \left( B \cap C \right) \cap D\)
(h)
\(\left( A \cap B \right) \cap \left( C \cap D \right)\)
3.
Based on the work in Exercise 2, does the placement of the parentheses matter when determining the union (or intersection) of these four sets? Does this make it possible to define \(A \cup B \cup C \cup D\) and \(A \cap B \cap C \cap D\text{?}\)
We have already seen that the elements of a set may themselves be sets. For example, the power set of a set \(T\text{,}\) \(\mathcal{P}( T )\text{,}\) is the set of all subsets of \(T\text{.}\) The phrase, “a set of sets” sounds confusing, and so we often use the terms collection and family when we wish to emphasize that the elements of a given set are themselves sets. We would then say that the power set of \(T\) is the family (or collection) of sets that are subsets of \(T\text{.}\)
One of the purposes of the work we have done so far in this activity was to show that it is possible to define the union and intersection of a family of sets.
Definition.
Let \(\mathcal{C}\) be a family of sets. The union of \(\mathbf{\mathcal{C}}\) is defined as the set of all elements that are in at least one of the sets in \(\mathcal{C}\text{.}\) We write
The intersection of \(\mathbf{\mathcal{C}}\) is defined as the set of all elements that are in all of the sets in \(\mathcal{C}\text{.}\) That is,
For example, consider the four sets \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\) used earlier in this activity and the sets
We can then consider the following families of sets: \(\mathscr{A} = \left\{A, B, C , D \right\}\) and \(\mathscr{B} = \left\{A, B, C , D, S, T \right\}\text{.}\)
4.
Explain why
and use your work in Exercise 1, Exercise 2, and Exercise 3 to determine \(\bigcup\limits_{X \in \mathscr{A}}^{}X\) and \(\bigcap\limits_{X \in \mathscr{A}}^{}X\text{.}\)
5.
Use the roster method to specify \(\bigcup\limits_{X \in \mathscr{B}}^{}X\) and \(\bigcap\limits_{X \in \mathscr{B}}^{}X\text{.}\)
6.
Use the roster method to specify the sets \(\left( \bigcup\limits_{X \in \mathscr{A}}^{}X \right)^c\) and \(\bigcap\limits_{X \in \mathscr{A}}^{}X^c\text{.}\) Remember that the universal set is \(\N\text{.}\)
Beginning Activity Beginning Activity 2: An Indexed Family of Sets
We often use subscripts to identify sets. For example, in Beginning Activity 1, instead of using \(A\text{,}\) \(B\text{,}\) \(C\text{,}\) and \(D\) as the names of the sets, we could have used \(A_1\text{,}\) \(A_2\text{,}\) \(A_3\text{,}\) and \(A_4\text{.}\) When we do this, we are using the subscript as an identifying tag, or index, for each set. We can also use this idea to specify an infinite family of sets. For example, for each natural number \(n\text{,}\) we define
So if we have a family of sets \(\mathscr{C} = \left\{ C_1, C_2, C_3, C_4 \right\}\text{,}\) we use the notation \(\bigcup\limits_{j=1}^{4}C_j\) to mean the same thing as \(\bigcup\limits_{X \in \mathscr{C}}^{}X\text{.}\)
1.
Determine \(\bigcup\limits_{j=1}^{4}C_j\) and \(\bigcap\limits_{j=1}^{4}C_j\text{.}\)
We can see that with the use of subscripts, we do not even have to define the family of sets \(\mathscr{A}\text{.}\) We can work with the infinite family of sets
and use the subscripts to indicate which sets to use in a union or an intersection.
2.
Use the roster method to specify each of the following pairs of sets. The universal set is \(\N\text{.}\)
(a)
\(\bigcup\limits_{j=1}^{6}C_j\) and \(\bigcap\limits_{j=1}^{6}C_j\)
(b)
\(\bigcup\limits_{j=1}^{8}C_j\) and \(\bigcap\limits_{j=1}^{8}C_j\)
(c)
\(\bigcup\limits_{j=4}^{8}C_j\) and \(\bigcap\limits_{j=4}^{8}C_j\)
(d)
\(\left( \bigcap\limits_{j=1}^{4}C_j \right)^c\) and \(\bigcup\limits_{j=1}^{4}C_j^c\)
Subsection The Union and Intersection of an Indexed Family of Sets
One of the purposes of the beginning activities was to show that we often encounter situations in which more than two sets are involved, and it is possible to define the union and intersection of more than two sets. In Beginning Activity 2, we also saw that it is often convenient to “index” the sets in a family of sets. In particular, if \(n\) is a natural number and \(\mathscr{A} = \left\{ A_1, A_2, \ldots , A_n \right\}\) is a family of \(n\) sets, then the union of these \(n\) sets, denoted by \(A_1 \cup A_2 \cup \cdots \cup A_n\) or \(\bigcup\limits_{j=1}^{n}A_j\text{,}\) is defined as
We can also define the intersection of these \(n\) sets, denoted by \(A_1 \cap A_2 \cap \cdots \cap A_n\) or \(\bigcap\limits_{j=1}^{n}A_j\text{,}\) as
We can also extend this idea to define the union and intersection of a family that consists of infinitely many sets. So if \(\mathscr{B} = \left\{ B_1, B_2, \ldots , B_n, \ldots \: \right\}\text{,}\) then
Progress Check 5.35. An Infinite Family of Sets.
For each natural number \(n\text{,}\) let \(A_n = \left\{ 1, n, n^2 \right\}\text{.}\) For example,
and
Determine each of the following sets:
(a)
\(\bigcup\limits_{j=1}^{6}A_j\)
\(\bigcup\limits_{j=1}^{6}A_j = \{1, 2, 3, 4, 5, 6, 9, 16, 25, 36 \}\)
(b)
\(\bigcap\limits_{j=1}^{6}A_j\)
\(\bigcap\limits_{j=1}^{6}A_j = \{ 1 \}\)
(c)
\(\bigcup\limits_{j=3}^{6}A_j\)
\(\bigcup\limits_{j=3}^{6}A_j = \{3, 4, 5, 6, 9, 16, 25, 36 \}\)
(d)
\(\bigcap\limits_{j=3}^{6}A_j\)
\(\bigcap\limits_{j=3}^{6}A_j = \{ 1 \}\)
(e)
\(\bigcup\limits_{j=1}^{\infty}A_j\)
\(\bigcup\limits_{j=1}^{\infty}A_j = \N\)
(f)
\(\bigcap\limits_{j=1}^{\infty}A_j\)
\(\bigcap\limits_{j=1}^{\infty}A_j = \{ 1 \}\)
In all of the examples we have studied so far, we have used \(\N\) or a subset of \(\N\) to index or label the sets in a family of sets. We can use other sets to index or label sets in a family of sets. For example, for each real number \(x\text{,}\) we can define \(B_x\) to be the closed interval \(\left[x, x + 2 \right]\text{.}\) That is,
So we make the following definition. In this definition, \(\Lambda\) is the uppercase Greek letter lambda and \(\alpha\) is the lowercase Greek letter alpha.
Definition.
Let \(\Lambda\) be a nonempty set and suppose that for each \(\alpha \in \Lambda\text{,}\) there is a corresponding set \(A_\alpha\text{.}\) The family of sets \(\left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) is called an indexed family of sets indexed by \(\Lambda\text{.}\) Each \(\alpha \in \Lambda\) is called an index and \(\Lambda\) is called an indexing set.
Progress Check 5.36. Indexed Families of Sets.
In each of the indexed families of sets that we seen so far, if the indices were different, then the sets were different. That is, if \(\Lambda\) is an indexing set for the family of sets \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\text{,}\) then if \(\alpha, \beta \in \Lambda\) and \(\alpha \ne \beta\text{,}\) then \(A_\alpha \ne A_\beta\text{.}\) (Note: The letter \(\beta\) is the Greek lowercase beta.)
(a)
Let \(\Lambda = \left\{ 1, 2, 3, 4 \right\}\text{,}\) and for each \(n \in \Lambda\text{,}\) let \(A_n = \left\{ 2n + 6, 16 - 2n \right\}\text{,}\) and let \(\mathscr{A} = \left\{A_1, A_2, A_3, A_4 \right\}\text{.}\) Determine \(A_1\text{,}\) \(A_2\text{,}\) \(A_3\text{,}\) and \(A_4\text{.}\)
\(A_1 = \{7, 14 \}, A_2 = \{10, 12 \}, A_3 = \{10, 12 \}, A_4 = \{8, 14 \}\text{.}\)
(b)
Is the following statement true or false for the indexed family \(\mathscr{A}\) in (1)?
For all \(m, n \in \Lambda\text{,}\) if \(m \ne n\text{,}\) then \(A_m \ne A_n\text{.}\)
The statement is false. For example, \(2 \ne 3\) and \(A_2 = A_3\text{.}\)
(c)
Now let \(\Lambda = \R\text{.}\) For each \(x \in \R\text{,}\) define \(B_x = \left\{ 0, x^2, x^4 \right\}\text{.}\) Is the following statement true for the indexed family of sets \(\mathscr{B} = \left\{ B_x \mid x \in \R \right\}\text{?}\)
For all \(x, y \in \R\text{,}\) if \(x \ne y\text{,}\) then \(B_x \ne B_y\text{.}\)
The statement is false. For example, \(1 \ne -1\) and \(B_1 = B_{-1}\text{.}\)
We now restate the definitions of the union and intersection of a family of sets for an indexed family of sets.
Definition.
Let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets. The union over \(\mathbf{\mathscr{A}}\) is defined as the set of all elements that are in at least one of sets \(A_\alpha\text{,}\) where \(\alpha \in \Lambda\text{.}\) We write
The intersection over \(\mathbf{\mathscr{A}}\) is the set of all elements that are in all of the sets \(A_\alpha\) for each \(\alpha \in \Lambda\text{.}\) That is,
Example 5.37. A Family of Sets Indexed by the Positive Real Numbers.
For each positive real number \(\alpha\text{,}\) let \(A_\alpha\) be the interval \(\left( -1, \alpha \right]\text{.}\) That is,
If we let \(\R^+\) be the set of positive real numbers, then we have a family of sets indexed by \(\R^+\text{.}\) We will first determine the union of this family of sets. Notice that for each \(\alpha \in \R^+\text{,}\) \(\alpha \in A_\alpha\text{,}\) and if \(y\) is a real number with \(-1 \lt y \leq 0\text{,}\) then \(y \in A_\alpha\text{.}\) Also notice that if \(y \in \R\) and \(y \lt -1\text{,}\) then for each \(\alpha \in \R^+\text{,}\) \(y \notin A_\alpha\text{.}\) With these observations, we conclude that
To determine the intersection of this family, notice that
if \(y \in \R\) and \(y \lt -1\text{,}\) then for each \(\alpha \in \R^+\text{,}\) \(y \notin A_\alpha\text{;}\)
if \(y \in \R\) and \(-1 \lt y \leq 0\text{,}\) then \(y \in A_\alpha\) for each \(\alpha \in \R^+\text{;}\) and
if \(y \in \R\) and \(y > 0\text{,}\) then if we let \(\beta = \dfrac{y}{2}\text{,}\) \(y > \beta\) and \(y \notin A_\beta\text{.}\)
From these observations, we conclude that
Progress Check 5.38. A Continuation of Example 5.37.
Using the family of sets from Example 5.37, for each \(\alpha \in \R^+\text{,}\) we see that
Use the results from Example 5.37 to help determine each of the following sets. For each set, use either interval notation or set builder notation.
(a)
\(\left( \bigcup\limits_{\alpha \in \R^+}^{}A_\alpha \right)^c\)
Since \(\bigcup\limits_{\alpha \in \R^+}^{}A_\alpha = \left( -1, \infty \right)\text{,}\) \(\left( \bigcup\limits_{\alpha \in \R^+}^{}A_\alpha \right)^c = (-\infty, 1 ]\text{.}\)
(b)
\(\bigcap\limits_{\alpha \in \R^+}^{}A_{\alpha}^c\)
\(\bigcap\limits_{\alpha \in \R^+}^{}A_{\alpha}^c = (-\infty, -1]\text{.}\)
(c)
\(\left( \bigcap\limits_{\alpha \in \R^+}^{}A_\alpha \right)^c\)
Since \(\bigcap\limits_{\alpha \in \R^+}^{}A_\alpha = \left( -1, 0 \right]\text{,}\) \(\left( \bigcap\limits_{\alpha \in \R^+}^{}A_\alpha \right)^c = (-\infty, -1] \cup (0, \infty)\text{.}\)
(d)
\(\bigcup\limits_{\alpha \in \R^+}^{}A_{\alpha}^c\)
\(\bigcup\limits_{\alpha \in \R^+}^{}A_{\alpha}^c = (-\infty, -1] \cup (0, \infty)\text{.}\)
Subsection Properties of Union and Intersection
In Theorem 5.39, we will prove some properties of set operations for indexed families of sets. Some of these properties are direct extensions of corresponding properties for two sets. For example, we have already proved De Morgan's Laws for two sets in Theorem 5.26. The work in the beginning activities and Progress Check 5.38 suggests that we should get similar results using set operations with an indexed family of sets. For example, in Beginning Activity 2, we saw that
Theorem 5.39.
Let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets, each of which is a subset of some universal set \(U\text{.}\) Then
For each \(\beta \in \Lambda\text{,}\) \(\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \subseteq A_\beta\text{.}\)
For each \(\beta \in \Lambda\text{,}\) \(A_\beta \subseteq \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\)
\(\displaystyle \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c = \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\)
\(\displaystyle \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c = \bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\)
Proof.
We will prove Item 1 and Item 3. The proofs of Item 2 and Item 4 are included in Exercise 4. So we let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets. To prove Item 1, we let \(\beta \in \Lambda\) and note that if \(x \in \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\text{,}\) then \(x \in A_\alpha\text{,}\) for all \(\alpha \in \Lambda\text{.}\) Since \(\beta\) is one element in \(\Lambda\text{,}\) we may conclude that \(x \in A_\beta\text{.}\) This proves that \(\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \subseteq A_\beta\text{.}\)
To prove Item 3, we will prove that each set is a subset of the other set. We first let \(x \in \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c\text{.}\) This means that \(x \notin \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \right)\text{,}\) and this means that
there exists a \(\beta \in \Lambda\) such that \(x \notin A_\beta\text{.}\)Hence, \(x \in A_{\beta}^c\text{,}\) which implies that \(x \in \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\text{.}\) Therefore, we have proved that
We now let \(y \in \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\text{.}\) This means that there exists a \(\beta \in \Lambda\) such that \(y \in A_{\beta}^c\) or \(y \notin A_\beta\text{.}\) However, since \(y \notin A_\beta\text{,}\) we may conclude that \(y \notin \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\) and, hence, \(y \in \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c\text{.}\) This proves that
Using the results in (53) and (54), we have proved that \(\left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c = \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\text{.}\)
Many of the other properties of set operations are also true for indexed families of sets. Theorem 5.40 states the distributive laws for set operations.
Theorem 5.40.
Let \(\Lambda\) be a nonempty indexing set, let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets, and let \(B\) be a set. Then
\(B \cap \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right)\text{,}\) and
\(B \cup \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcap\limits_{\alpha \in \Lambda}^{} \left( B \cup A_{\alpha} \right)\text{.}\)
The proof of Theorem 5.40 is Exercise 5.
Subsection Pairwise Disjoint Families of Sets
In Section 5.2, we defined two sets \(A\) and \(B\) to be disjoint provided that \(A \cap B = \emptyset\text{.}\) In a similar manner, if \(\Lambda\) is a nonempty indexing set and \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) is an indexed family of sets, we can say that this indexed family of sets is disjoint provided that \(\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha = \emptyset\text{.}\) However, we can use the concept of two disjoint sets to define a somewhat more interesting type of “disjointness” for an indexed family of sets.
Definition.
Let \(\Lambda\) be a nonempty indexing set, and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets. We say that \(\mathscr{A}\) is pairwise disjoint provided that for all \(\alpha\) and \(\beta\) in \(\Lambda\text{,}\) if \(A_\alpha \ne A_\beta\text{,}\) then \(A_\alpha \cap A_\beta = \emptyset\text{.}\)
Progress Check 5.41. Disjoint Families of Sets.
Figure 5.42 shows two families of sets,
(a)
Is the family of sets \(\mathscr{A}\) a disjoint family of sets? A pairwise disjoint family of sets?
(b)
Is the family of sets \(\mathscr{B}\) a disjoint family of sets? A pairwise disjoint family of sets?
(c)
Now let the universal set be \(\R\text{.}\) For each \(n \in \N\text{,}\) let \(C_n = \left( n, \infty \right)\text{,}\) and let \(\mathscr{C} = \left\{ C_n \mid n \in \N \right\}\text{.}\)
Is the family of sets \(\mathscr{C}\) a disjoint family of sets? A pairwise disjoint family of sets?
All three families of sets \(\left( \mathscr{A}, \mathscr{B}, \text{ and } \mathscr{C} \right)\) are disjoint families of sets. Only the family \(\mathscr{A}\) is a pairwise disjoint family of sets.
Exercises Exercises
1.
For each natural number \(n\text{,}\) let \(A_n = \left\{n, n+1, n+2, n+3 \right\}\text{.}\) Use the roster method to specify each of the following sets:
(a)
\(\bigcap\limits_{j=1}^{3}A_j\)
\(\left\{ 3, 4 \right\}\)
(b)
\(\bigcup\limits_{j=1}^{3}A_j\)
(c)
\(\bigcap\limits_{j=3}^{7}A_j\)
(d)
\(\bigcup\limits_{j=3}^{7}A_j\)
\(\left\{ 3, 4, 5, 6, 7, 8, 9, 10 \right\}\)
(e)
\(A_9 \cap \left(\:\bigcup\limits_{j=3}^{7}A_j \right)\)
(f)
\(\bigcup\limits_{j=3}^{7}\left( A_9 \cap A_j \right)\)
2.
For each natural number \(n\text{,}\) let \(A_n = \left\{ k \in \N \left| \hspace{3pt} k \geq n \right. \right\}\text{.}\) Assuming the universal set is \(\N\text{,}\) use the roster method or set builder notation to specify each of the following sets:
(a)
\(\bigcap\limits_{j=1}^{5}A_j\)
\(\left\{ 5, 6, 7, \ldots \,\right\}\)
(b)
\(\left(\: \bigcap\limits_{j=1}^{5}A_j \right)^c\)
(c)
\(\bigcap\limits_{j=1}^{5}A_j^c\)
\(\emptyset\)
(d)
\(\bigcup\limits_{j=1}^{5}A_j^c\)
\(\left\{1, 2, 3, 4 \right\}\)
(e)
\(\bigcup\limits_{j=1}^{5}A_j\)
(f)
\(\left(\: \bigcup\limits_{j=1}^{5}A_j \right)^c\)
\(\emptyset\)
(g)
\(\bigcap\limits_{j \in \N}^{}A_j\)
(h)
\(\bigcup\limits_{j \in \N}^{}A_j\)
3.
For each positive real number \(r\text{,}\) define \(T_r\) to be the closed interval \(\left[ -r^2, r^2 \right]\text{.}\) That is,
Let \(\Lambda = \left\{ m \in \N \left| 1 \leq m \leq 10 \right. \right\}\text{.}\) Use either interval notation or set builder notation to specify each of the following sets:
(a)
\(\bigcup\limits_{k \in \Lambda}^{}T_k\)
\(\left\{ x \in \R \left| -100 \leq x \leq 100 \right. \right\}\)
(b)
\(\bigcap\limits_{k \in \Lambda}^{}T_k\)
\(\left\{ x \in \R \left| -1 \leq x \leq 1 \right. \right\}\)
(c)
\(\bigcup\limits_{r \in \R^+}^{}T_r\)
(d)
\(\bigcap\limits_{r \in \R^+}^{}T_r\)
(e)
\(\bigcup\limits_{k \in \N}^{}T_k\)
(f)
\(\bigcap\limits_{k \in \N}^{}T_k\)
4.
Prove Item 2 and Item 4 of Theorem 5.39. Let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets.
(a)
For each \(\beta \in \Lambda\text{,}\) \(A_\beta \subseteq \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\)
We let \(\beta \in \Lambda\) and let \(x \in A_\beta\text{.}\) Then \(x \in A_\alpha\text{,}\) for at least one \(\alpha \in \Lambda\) and, hence, \(x \in \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\) This proves that \(A_\beta \subseteq \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\)
(b)
\(\left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha \right)^c = \bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha}^c\)
5.
Prove Theorem 5.40. Let \(\Lambda\) be a nonempty indexing set, let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets, and let \(B\) be a set. Then
(a)
\(B \cap \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right)\text{,}\) and
We first let \(x \in B \cap \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right)\text{.}\) Then \(x \in B\) and \(x \in \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}\text{.}\) This means that there exists an \(\alpha \in \Lambda\) such that \(x \in A_\alpha\text{.}\) Hence, \(x \in B \cap A_\alpha\text{,}\) which implies that \(x \in \bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right)\text{.}\) This proves that \(B \cap \left(\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) \subseteq \bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right)\text{.}\)
We now let \(y \in \bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right)\text{.}\) So there exists an \(\alpha \in \Lambda\) such that \(y \in B \cap A_{\alpha}\text{.}\) Then \(y \in B\) and \(y \in A_{\alpha}\text{,}\) which implies that \(y \in B\) and \(y \in \bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha}\text{.}\) Therefore, \(y \in B \cap \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right)\text{,}\) and this proves that \(\bigcup\limits_{\alpha \in \Lambda}^{} \left( B \cap A_{\alpha} \right) \subseteq B \cap \left(\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right)\text{.}\)
(b)
\(B \cup \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcap\limits_{\alpha \in \Lambda}^{} \left( B \cup A_{\alpha} \right)\text{.}\)
6.
Let \(\Lambda\) and \(\Gamma\) be nonempty indexing sets. (Note: The letter \(\Gamma\) is the uppercase Greek letter gamma.) Also, let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) and \(\mathscr{B} = \left\{ B_\beta \mid \beta \in \Gamma \right\}\) be indexed families of sets. Use the distributive laws in Exercise 5 to:
(a)
Write \(\left(\: \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\right) \cap \left(\: \bigcup\limits_{\beta \in \Gamma}^{}B_\beta \right)\) as a union of intersections of two sets.
(b)
Write \(\left(\: \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\right) \cup \left(\: \bigcap\limits_{\beta \in \Gamma}^{}B_\beta \right)\) as an intersection of unions of two sets.
7.
Let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets. Also, assume that \(\Gamma \subseteq \Lambda\) and \(\Gamma \ne \emptyset\text{.}\) Prove that
(a)
\(\bigcup\limits_{\alpha \in \Gamma}^{}A_\alpha \subseteq \bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha\)
(b)
\(\bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha \subseteq \bigcap\limits_{\alpha \in \Gamma}^{}A_\alpha\)
8.
Let \(\Lambda\) be a nonempty indexing set and let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets.
(a)
Prove that if \(B\) is a set such that \(B \subseteq A_\alpha\) for every \(\alpha \in \Lambda\text{,}\) then \(B \subseteq \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\)
Let \(x \in B\text{.}\) For each \(\alpha \in \Lambda\text{,}\) \(B \subseteq A_\alpha\) and, hence, \(x \in A_\alpha\text{.}\) This means that for each \(\alpha \in \Lambda\text{,}\) \(x \in A_\alpha\) and, hence, \(x \in \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\) Therefore, \(B \subseteq \bigcap\limits_{\alpha \in \Lambda}^{}A_\alpha\text{.}\)
(b)
Prove that if \(C\) is a set such that \(A_\alpha \subseteq C\) for every \(\alpha \in \Lambda\text{,}\) then \(\bigcup\limits_{\alpha \in \Lambda}^{}A_\alpha \subseteq C\text{.}\)
9.
For each natural number \(n\text{,}\) let \(A_n = \left\{ x \in \R \left| \hspace{3pt} n -1 \lt x \lt n \right. \right\}\text{.}\) Prove that \(\left\{ A_n \left| \hspace{3pt} n \in \N \right. \right\}\) is a pairwise disjoint family of sets and that \(\bigcup\limits_{n \in \N}^{}A_n = \left( \R^+ - \N \right)\text{.}\)
10.
For each natural number \(n\text{,}\) let \(A_n = \left\{ k \in \N \left| \hspace{3pt} k \geq n \right. \right\}\text{.}\) Determine if the following statements are true or false. Justify each conclusion.
(a)
For all \(j, k \in \N\text{,}\) if \(j \ne k\text{,}\) then \(A_j \cap A_k \ne \emptyset\text{.}\)
(b)
\(\bigcap\limits_{k \in \N}^{}A_k = \emptyset\)
11.
Give an example of an indexed family of sets \(\left\{ A_n \left| n \in \N \right. \right\}\) such all three of the following conditions are true:
For each \(m \in \N\text{,}\) \(A_m \subseteq (0, 1)\text{;}\)
For each \(j, k \in \N\text{,}\) if \(j \ne k\text{,}\) then \(A_j \cap A_k \ne \emptyset\text{;}\) and
\(\bigcap\limits_{k \in \N}^{}A_k = \emptyset\text{.}\)
12.
Let \(\Lambda\) be a nonempty indexing set, let \(\mathscr{A} = \left\{ A_\alpha \mid \alpha \in \Lambda \right\}\) be an indexed family of sets, and let \(B\) be a set. Use the results of Theorem 5.39 and Theorem 5.40 to prove each of the following:
(a)
\(\left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) - B = \bigcup\limits_{\alpha \in \Lambda}^{} \left( A_{\alpha} - B \right)\)
We first rewrite the set difference and then use a distributive law.
(b)
\(\left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) - B = \bigcap\limits_{\alpha \in \Lambda}^{} \left( A_{\alpha} - B \right)\)
(c)
\(B - \left(\:\bigcup\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcap\limits_{\alpha \in \Lambda}^{} \left( B - A_{\alpha} \right)\)
(d)
\(B - \left(\:\bigcap\limits_{\alpha \in \Lambda}^{}A_{\alpha} \right) = \bigcup\limits_{\alpha \in \Lambda}^{} \left( B - A_{\alpha} \right)\)
Activity 33. An Indexed Family of Subsets of the Cartesian Plane.
Let \(\R^*\) be the set of nonnegative real numbers, and for each \(r \in \R^*\text{,}\) let
If \(r > 0\text{,}\) then the set \(C_r\) is the circle of radius \(r\) with center at the origin as shown in Figure 5.43, and the set \(D_r\) is the shaded disk (including the boundary) shown in Figure 5.43. 
(a)
Determine \(\bigcup\limits_{r \in \R^*}^{}C_r\) and \(\bigcap\limits_{r \in \R^*}^{}C_r\text{.}\)
(b)
Determine \(\bigcup\limits_{r \in \R^*}^{}D_r\) and \(\bigcap\limits_{r \in \R^*}^{}D_r\text{.}\)
(c)
Determine \(\bigcup\limits_{r \in \R^*}^{}T_r\) and \(\bigcap\limits_{r \in \R^*}^{}T_r\text{.}\)
(d)
Let \(\mathscr{C} = \left\{ C_r \mid r \in \R^* \right\}\text{,}\) \(\mathscr{D} = \left\{ D_r \mid r \in \R^* \right\}\text{,}\) and \(\mathscr{T} = \left\{ T_r \mid r \in \R^* \right\}\text{.}\) Are any of these indexed families of sets pairwise disjoint? Explain.
Now let \(I\) be the closed interval \([0, 2]\) and let \(J\) be the closed interval \([1, 2]\text{.}\)
(e)
Determine \(\bigcup\limits_{r \in I}^{}C_r\text{,}\) \(\bigcap\limits_{r \in I}^{}C_r\text{,}\) \(\bigcup\limits_{r \in J}^{}C_r\text{,}\) and \(\bigcap\limits_{r \in J}^{}C_r\text{.}\)
(f)
Determine \(\bigcup\limits_{r \in I}^{}D_r\text{,}\) \(\bigcap\limits_{r \in I}^{}D_r\text{,}\) \(\bigcup\limits_{r \in J}^{}D_r\text{,}\) and \(\bigcap\limits_{r \in J}^{}D_r\text{.}\)
(g)
Determine \(\left(\: \bigcup\limits_{r \in I}^{}D_r \right)^c\text{,}\) \(\left(\: \bigcap\limits_{r \in I}^{}D_r \right)^c\text{,}\) \(\left(\: \bigcup\limits_{r \in J}^{}D_r \right)^c\text{,}\) and \(\left(\: \bigcap\limits_{r \in J}^{}D_r \right)^c\text{.}\)
(h)
Determine \(\bigcup\limits_{r \in I}^{}T_r\text{,}\) \(\bigcap\limits_{r \in I}^{}T_r\text{,}\) \(\bigcup\limits_{r \in J}^{}T_r\text{,}\) and \(\bigcap\limits_{r \in J}^{}T_r\text{.}\)
(i)
Use De Morgan's Laws to explain the relationship between your answers in Task 33.g and Task 33.h.