Question

Prove that if A and B are any two sets such that A is proper subset of B, then A union B=B-
1) by direct method
2) by proving its contrapositive.
3) by contradiction.

Answer 1

Answer on Question #47046 – Math - Set Theory

Problem.

Prove that if A and B are any two sets such that A is proper subset of B, then A union B=B-

1) by direct method

2) by proving its contrapositive.

3) by contradiction.

Solution.

Consider 2 sets $A, B$.

As $A \subset B$ then $\forall x [x \in A \Rightarrow x \in B]$.

As $A$ is a proper subset of $B$ then $\exists x [x \in B \land x \notin A]$.

1) by direct method

$P \Rightarrow Q$.

$\forall x [x \in A \cup B \Rightarrow x \in A \lor x \in B \Rightarrow x \in B \lor x \in B \Rightarrow x \in B]$.

Vice versa, $\forall x [x \in B \Rightarrow x \in A \lor x \in B \Rightarrow x \in A \cup B]$. The proof is complete.

2) by proving its contrapositive $\neg Q \Rightarrow \neg P$.

Obviously, $A \cup B \supseteq B$. Therefore, $A \cup B \neq B \Rightarrow \exists x' [x' \in A \cup B \land x' \notin B]$.

So, $\exists x' [x' \in A \cup B \land x' \notin B \Rightarrow (x' \in A \lor x' \in B) \land x' \notin B \Rightarrow (x' \in A \land x' \notin B) \lor (x' \in B \land x' \notin B) \Rightarrow (x' \in A \land x' \notin B)]$, which is the negation to $\forall x [x \in A \Rightarrow x \in B]$-in other words, the negation to $A \subset B$-and the proof is complete.

3) by contradiction $P \land \neg Q$.

To prove the proposition given, we consider the opposite $-P \land \neg Q$. Remember that in the second part we proved that $\neg Q \Rightarrow \neg P$. That's why $P \land \neg Q \Rightarrow P \land \neg P \equiv \bot$ which arrives at the contradiction we want. The proof is complete.

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