MAT102 Lecture 04B

02SetsAndPropositionsExercises_Student.pdf

• Suppose $A,B,C$ are sets

• Show that $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$

• Double Subset Inclusion

• $\subseteq$

  • Let $x\in A\cap (B\cup C)$
  • So $x\in A$ and $x\in B$ or $x\in C$
  • Method 1: Cases
    • Case 1: Suppose $x\in B$
      • Since $x\in A$ then $x\in A\cap B$
      • Notice $(A\cap B)\subseteq (A\cap B)\cup (A\cap C)$
    • Case 2: Suppose $x\in C$
      • Since $x\in A$ then $x\in A\cap C$
      • So $(A\cap C)\subseteq (A\cap B)\cup (A\cap C)$
  • Method 2: Direct
    • Without loss of generality, suppose $x\in B$
      • The reason this works, is because $B$ vs $C$ doesn't matter. Because we have $B\cup C$ so either way, you can see by the cases approach, we have the same exact thing.
    • If $x\in C$ then relabel that set as $B$
    • So if $x\in A$ and $x\in B$
    • Then $x$ must be in $x\in (A\cap B)\subseteq (A\cap B)\cup (A\cap C)$
    • So we're done

• $\supseteq$

  • Let $x\in (A\cap B)\cup (A\cap C)$
  • So $x\in A\cap B$ or $x\in A\cap C$
  • Method 1: Cases
    • Case 1: Suppose $x\in A\cap B$
      • Notice $x\in A\cap B$ so $A\cap B\subseteq A\cap (B\cup C)$
      • As $C$ has no bearing on this.
    • Case 2: Suppose $x\in A\cap C$
      • Notice that $x\in A\cap C$ so $A\cap C\subseteq A\cap (B\cup C)$
  • Method 2: Direct
    • If $x\in A\cap B$ or $x\in A\cap C$
    • Then $x\in A\cap (B\cup C)$ as required

• $◻$

• 1

  • Define a new operator $\otimes$ as $A\otimes B$ as $\mathrm{\neg }(A\wedge B)$
  • A
    • Show $A\otimes A$ and see that $\mathrm{\neg }A=A\otimes A$
    • $$\begin{array}{cccc}A& A& A\wedge A& \mathrm{\neg }(A\wedge B)\\ T& T& T& F\\ F& F& F& T\end{array}$$
  • B
    • $$\begin{array}{ccccc}A& B& (A\wedge B)& \mathrm{\neg }(A\wedge B)& \mathrm{\neg }(\mathrm{\neg }(A\wedge B))\wedge (\mathrm{\neg }(A\wedge B))\\ T& T& T& F& T\\ T& F& F& T& F\\ F& T& F& T& F\\ F& F& F& T& F\end{array}$$
  • C
    • Find out how to write $A\vee B$ as $A,B$ and $\otimes$
    • $$\begin{array}{cccccc}A& B& A\vee B& (A\otimes A)& (B\otimes B)& (A\otimes A)\otimes (B\otimes B)\\ T& T& T& F& F& T\\ T& F& T& F& T& T\\ F& T& T& T& F& T\\ F& F& F& T& T& F\end{array}$$
    • Notice that $A\vee B=(A\otimes A)\otimes (B\otimes B)$