Answer to Question #126707 in Discrete Mathematics for kavee

Question

10. Let A , B, and C be sets. Show that(a) (A ∪ B) ⊆ (A ∪ B ∪ C)
(b) (A ∩ B ∩ C) ⊆ (A ∩ B)
(c) (A − B) − C ⊆ (A − C)
(d) (A − C) ∩ (C − B) = ∅
(e) (B − A) ∪ (C − B) = ∅

Answer 1

(a) Let us take an arbitrary $x\in A\cup B$ . We have: $x\in A\cup B \Leftrightarrow (x\in A) \lor (x\in B)\Rightarrow (x\in A) \lor (x\in B)\lor (x\in C)\Leftrightarrow x\in A\cup B\cup C$

Thus, we receive that $(A\cup B)\subseteq (A\cup B\cup C)$

(b) We take an arbitrary $x\in A\cap B\cap C$ . By definition, we have

$x\in A\cap B\cap C\Leftrightarrow x\in A \land x\in B \land x\in C\Rightarrow x\in A \land x\in B\Leftrightarrow$

$\Leftrightarrow x\in A\cap B$ . Thus, we receive $A\cap B\cap C\subseteq A\cap B$ .

(c) We assume that the sign "-" denotes the subtraction of sets. Then for an arbitrary $x\in(A-B)-C$ we have:

$x\in(A-B)-C \Leftrightarrow x\in(A-B)\land x\notin C \Leftrightarrow x\in A \land x\notin B \land x\notin C \Rightarrow x\in A \land x\notin C \Rightarrow x\in (A-C)$

Thus, we got: $(A-B)-C\subseteq (A-C)$

(d) For an arbitrary $x\in (A-C)\cap(C-B)$ we have:

$x\in (A-C)\cap(C-B)\Leftrightarrow x\in (A-C) \land x \in (C-B) \Leftrightarrow$

$x\in A \land x\notin C \land x\in C \land x\notin B$ . The latter means that the set $(A-C)\cap(C-B)$ is empty. Thus, $(A-C)\cap(C-B)=\varnothing$

(e) Let us provide a counterexample: $B=C=\{1\}, A=\varnothing$ .

I.e., B and C consist of one natural number and set A is empty. Then, $(B-A)\cup(C-B)=\{1\}$

Thus, the statement is false.