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

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Let $A, B,$ and $C$ be sets.

a) Let us show that $(A ∪ B) ⊆ (A ∪ B ∪ C).$

Let $x\in A ∪ B$. Then $x\in A$ or $x\in B$. It follows that $x\in A$ or $x\in B$ or $x\in C$, and hence $x\in A ∪ B∪ C.$ We conclude that $(A ∪ B) ⊆ (A ∪ B ∪ C).$

b) Let us show that $(A ∩ B ∩ C) ⊆ (A ∩ B).$

Let $x\in A \cap B \cap C$. Then $x\in A$ and $x\in B$ and $x\in C$. It follows that $x\in A$ and $x\in B$, and hence $x\in A \cap B.$ We conclude that $(A ∩ B ∩ C) ⊆ (A ∩ B).$

c) Let us show that $(B − A) ∪ (C − A) = (B ∪ C) − A.$

Taking into account that $X-Y=X\cap \overline{Y}$ and using distributive law, we conclude that

$(B − A) ∪ (C − A) =(B\cap\overline{A})\cup(C\cap\overline{A}) =(B\cup C)\cap\overline{A} = (B ∪ C) − A.$