Prove that the three statements are equivalent

Question

I need to show that the following statements are equivalent.

A $\subset$ B, A $\cap$ B$^c$ = $\emptyset$, and A$^c \cup$ B = U (U is the universal set)

So to show that A $\subset$ B is true I said if x$\in$A and and if x$\in$B, then A $\subset$ B is true.

For A $\cap$ B$^c$ = $\emptyset$, I said, if x$\in$A and x$\notin$B, then the intersection is empty so this is true.

I am interpreting the previous statement as, x is an element in A and x is not an element in B therefore, the intersection of A and B is empty. Is that correct?

If I am reading this properly, I cannot see how A $\subset$ B implies A $\cap$ B$^c$ = $\emptyset$, and this is what I need help with.

Thanks,

Tony

Answer 1

You want to show that A ⊂ B implies A ∩ (B complement) = ∅ and vice versa.

Suppose that A ⊂ B. This means that for all a in A we have that a is in B. So is it possible for there to be an a' in A that is not in B?

If it is not possible then we have shown that none of the a's in A are in B complement. So what does that mean about their intersection.

Think about it.

Answer 2

We are given three statements: $$ \begin{align} \text{(a) }&A \subset B\\ \text{(b) }&A \cap B^c = \phi\\ \text{(c) }&A^c \cup B = U \end{align} $$ To prove that the three statements are equivalent we show that $\text{(a)}\implies \text{(b)},\text{(b)}\implies\text{(c)},\text{ and }\text{(c)}\implies \text{(a)}$

1. prove $\text{(a)}\implies \text{(b)}$: $$ \begin{align} & A \subset B\\ \implies & x\in B\ \forall x \in A \\ \implies & \{x: x\in A, x \notin B \} = \phi\\ \implies & A \cap B^c = \phi \end{align} $$
2. prove $\text{(b)}\implies \text{(c)}$: $$ \begin{align} &A\cap B^c = \phi\\ \implies &\left(A \cap B^c\right)^c = \phi^c\\ \implies &A^c\cup B = U \end{align} $$
3. prove $\text{(c)}\implies \text{(a)}$: $$ \begin{align} &A^c\cup B = U\\ \implies &\{x:x \notin A\text{ or }x\in B\} = U\\ \implies &\{x:x \in A \text{ and } x\notin B\} = \phi\\ \implies &x\in B\ \forall x \in A \\ \implies &A \subset B \end{align} $$

Hence the equivalence of the three statements is proved.