Mathematical-Reasoning · Unit 19 · Video 4 · Interactive Practice

De Morgan's Laws for Any Family, and Pairwise Disjoint

IKey Formulas

FormulaNameKey step
⋂α∈ΛAα⊆Aβ⊆⋃α∈ΛAα\bigcap_{\alpha\in\Lambda} A_\alpha \subseteq A_\beta \subseteq \bigcup_{\alpha\in\Lambda} A_\alphaEvery set lies between ⋂\bigcap and ⋃\bigcupβ\beta is one of the α\alpha's, and β\beta is a witness for ∃\exists
(⋂α∈ΛAα)c=⋃α∈ΛAαc\left(\bigcap_{\alpha\in\Lambda} A_\alpha\right)^c = \bigcup_{\alpha\in\Lambda} A_\alpha^c
(⋃α∈ΛAα)c=⋂α∈ΛAαc\left(\bigcup_{\alpha\in\Lambda} A_\alpha\right)^c = \bigcap_{\alpha\in\Lambda} A_\alpha^c		De Morgan's laws¬∀⇔∃¬\neg\forall \Leftrightarrow \exists\neg and ¬∃⇔∀¬\neg\exists \Leftrightarrow \forall\neg
B∩(⋃α∈ΛAα)=⋃α∈Λ(B∩Aα)B \cap \left(\bigcup_{\alpha\in\Lambda} A_\alpha\right) = \bigcup_{\alpha\in\Lambda} (B \cap A_\alpha)
B∪(⋂α∈ΛAα)=⋂α∈Λ(B∪Aα)B \cup \left(\bigcap_{\alpha\in\Lambda} A_\alpha\right) = \bigcap_{\alpha\in\Lambda} (B \cup A_\alpha)		Distributive lawsLaw (2) splits on y∈By \in B or y∉By \notin B
Disjoint: ⋂α∈ΛAα=∅\bigcap_{\alpha\in\Lambda} A_\alpha = \emptyset
Pairwise: Aα≠Aβ⇒Aα∩Aβ=∅A_\alpha \ne A_\beta \Rightarrow A_\alpha \cap A_\beta = \emptyset		Two kinds of disjointPairwise disjoint ⇒\Rightarrow disjoint when the family has at least two different sets

Key Insight: Negation exchanges ∀\forall and ∃\exists, so a complement exchanges ⋂\bigcap and ⋃\bigcup. The statement x∉⋂α∈ΛAαx \notin \bigcap_{\alpha\in\Lambda} A_\alpha says xx misses some AβA_\beta, not that xx misses every AαA_\alpha.

IIThe Complement Swaps ⋂\bigcap and ⋃\bigcup

An element outside the intersection misses at least one set; an element outside the union misses every set.

IIIDisjoint versus Pairwise Disjoint

A family is disjoint when no point lies in every set, and pairwise disjoint when no two different sets overlap.

💡 Challenge: stack all four discs on one spot. The family then has only one set, so it is pairwise disjoint with no pair to check, yet its intersection is that set, which is not empty.

IVNested Rays Cn=(n,∞)C_n = (n, \infty)

Any two of the sets overlap, since Cm∩Cn=Cmax⁡(m,n)C_m \cap C_n = C_{\max(m,n)}, yet no real number lies in every CnC_n.

VQuiz Questions

Problem 1 · Negating the Intersection

Given: Λ≠∅\Lambda \ne \emptyset and a family {Aα∣α∈Λ}\{A_\alpha \mid \alpha \in \Lambda\}. Which statement is equivalent to x∉⋂α∈ΛAαx \notin \bigcap_{\alpha\in\Lambda} A_\alpha?

✅ Correct! Being in the intersection is a "for all" statement, and its negation is "there exists an index where it fails".
❌ That is the common slip. "xx is in no AαA_\alpha" is much stronger: an element can miss one set and still lie in all the others. This statement says x∉⋃α∈ΛAαx \notin \bigcup_{\alpha\in\Lambda} A_\alpha.
❌ Not quite. This says x∈⋃α∈ΛAαx \in \bigcup_{\alpha\in\Lambda} A_\alpha. Negating "∀α, x∈Aα\forall \alpha,\ x \in A_\alpha" negates the membership too: x∉Aβx \notin A_\beta.
❌ Not quite. This is the definition of x∈⋂α∈ΛAαx \in \bigcap_{\alpha\in\Lambda} A_\alpha itself, not its negation.
Show solution

By the definition of intersection, x∈⋂α∈ΛAα  ⟺  ∀α∈Λ, x∈Aαx \in \bigcap_{\alpha\in\Lambda} A_\alpha \iff \forall \alpha \in \Lambda,\ x \in A_\alpha. Negate both sides:

x∉⋂α∈ΛAα  ⟺  ¬(∀α∈Λ, x∈Aα)  ⟺  ∃β∈Λ, x∉Aβ.x \notin \bigcap_{\alpha\in\Lambda} A_\alpha \iff \neg\big(\forall \alpha \in \Lambda,\ x \in A_\alpha\big) \iff \exists \beta \in \Lambda,\ x \notin A_\beta.

The last statement says x∈Aβcx \in A_\beta^c for some β\beta, that is, x∈⋃α∈ΛAαcx \in \bigcup_{\alpha\in\Lambda} A_\alpha^c. This is the first inclusion of De Morgan's law (3).

Problem 2 · An Infinite Family

Given: universal set R\mathbb{R} and Aα=(−1,α]A_\alpha = (-1, \alpha] for each α∈R+\alpha \in \mathbb{R}^+. Find (⋂α∈R+Aα)c\left(\bigcap_{\alpha\in\mathbb{R}^+} A_\alpha\right)^c.

✅ Correct! The intersection is (−1,0](-1, 0], and its complement equals ⋃α∈R+Aαc\bigcup_{\alpha\in\mathbb{R}^+} A_\alpha^c.
❌ Check the endpoints. 0∈(−1,α]0 \in (-1, \alpha] for every α>0\alpha > 0, so 00 is in the intersection and not in its complement; −1-1 is in no AαA_\alpha, so it is in the complement.
❌ That is the complement of the union. ⋃αAα=(−1,∞)\bigcup_{\alpha} A_\alpha = (-1, \infty), so (−∞,−1](-\infty, -1] is (⋃αAα)c=⋂αAαc\left(\bigcup_\alpha A_\alpha\right)^c = \bigcap_\alpha A_\alpha^c: the points that miss every set. The question asks for points that miss at least one set.
❌ Not quite. A number x≤−1x \le -1 is in no AαA_\alpha at all, so it is certainly outside the intersection.
Show solution

The intersection. If −1<x≤0-1 \lt x \le 0, then x≤0<αx \le 0 \lt \alpha for every α∈R+\alpha \in \mathbb{R}^+, so xx is in every AαA_\alpha. If x>0x \gt 0, the index β=x/2\beta = x/2 gives x∉(−1,β]x \notin (-1, \beta]. If x≤−1x \le -1, xx is in no AαA_\alpha. Hence ⋂α∈R+Aα=(−1,0]\bigcap_{\alpha\in\mathbb{R}^+} A_\alpha = (-1, 0].

(⋂α∈R+Aα)c=(−1,0]c=(−∞,−1]∪(0,∞).\Big(\bigcap_{\alpha\in\mathbb{R}^+} A_\alpha\Big)^c = (-1, 0]^c = (-\infty, -1] \cup (0, \infty).

Check with De Morgan. Aαc=(−∞,−1]∪(α,∞)A_\alpha^c = (-\infty, -1] \cup (\alpha, \infty), and the union of these over all α>0\alpha \gt 0 is (−∞,−1]∪(0,∞)(-\infty, -1] \cup (0, \infty), the same set.

Problem 3 · Overlapping Neighbours

Given: universal set R\mathbb{R} and Dn=[n,n+2)D_n = [n, n+2) for each n∈Nn \in \mathbb{N}, with D={Dn∣n∈N}\mathcal{D} = \{D_n \mid n \in \mathbb{N}\}. Which description of D\mathcal{D} is correct?

✅ Correct! Neighbouring sets overlap, but D1∩D3=∅D_1 \cap D_3 = \emptyset already forces the whole intersection to be empty.
❌ Check neighbours. D1∩D2=[2,3)≠∅D_1 \cap D_2 = [2, 3) \ne \emptyset, so two different sets of D\mathcal{D} overlap.
❌ That cannot happen here. D\mathcal{D} has at least two different sets, and then pairwise disjoint implies disjoint. Besides, D1∩D2=[2,3)≠∅D_1 \cap D_2 = [2, 3) \ne \emptyset.
❌ Not quite. Overlapping neighbours do not give a point in every set: D1∩D3=[1,3)∩[3,5)=∅D_1 \cap D_3 = [1, 3) \cap [3, 5) = \emptyset.
Show solution

Not pairwise disjoint. D1=[1,3)D_1 = [1, 3) and D2=[2,4)D_2 = [2, 4) are different sets with D1∩D2=[2,3)≠∅D_1 \cap D_2 = [2, 3) \ne \emptyset.

Disjoint. By part (1) of the theorem, ⋂n∈NDn⊆D1\bigcap_{n\in\mathbb{N}} D_n \subseteq D_1 and ⋂n∈NDn⊆D3\bigcap_{n\in\mathbb{N}} D_n \subseteq D_3, so

⋂n∈NDn⊆D1∩D3=[1,3)∩[3,5)=∅.\bigcap_{n\in\mathbb{N}} D_n \subseteq D_1 \cap D_3 = [1, 3) \cap [3, 5) = \emptyset.

So D\mathcal{D} is disjoint but not pairwise disjoint, like the families B\mathcal{B} and C\mathcal{C} in the video.

Problem 4 · A Repeated Set

Given: Λ={1,2}\Lambda = \{1, 2\} with A1={7}A_1 = \{7\} and A2={7}A_2 = \{7\}. Which description of the family {Aα∣α∈Λ}\{A_\alpha \mid \alpha \in \Lambda\} is correct?

✅ Correct! The family has only one set, {7}\{7\}, so there is no pair of different sets to check, while ⋂α∈ΛAα={7}≠∅\bigcap_{\alpha\in\Lambda} A_\alpha = \{7\} \ne \emptyset.
❌ Read the condition again. Pairwise disjoint only tests Aα∩AβA_\alpha \cap A_\beta when Aα≠AβA_\alpha \ne A_\beta. Here A1=A2A_1 = A_2, so the overlap {7}\{7\} is never tested.
❌ Not quite. A1∩A2={7}A_1 \cap A_2 = \{7\} is not empty, so the family is not disjoint.
❌ Not quite. The intersection A1∩A2={7}A_1 \cap A_2 = \{7\} is not empty, so the family is not disjoint.
Show solution

Pairwise disjoint. The condition is ∀α,β∈Λ, Aα≠Aβ⇒Aα∩Aβ=∅\forall \alpha, \beta \in \Lambda,\ A_\alpha \ne A_\beta \Rightarrow A_\alpha \cap A_\beta = \emptyset. Since A1=A2A_1 = A_2, every hypothesis Aα≠AβA_\alpha \ne A_\beta is false, so every implication is true.

Not disjoint. ⋂α∈ΛAα={7}∩{7}={7}≠∅\bigcap_{\alpha\in\Lambda} A_\alpha = \{7\} \cap \{7\} = \{7\} \ne \emptyset.

This does not contradict "pairwise disjoint implies disjoint": that result needs at least two different sets, and this family is {{7}}\{\{7\}\}, with exactly one.

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