(Dis-)proving statements about sets

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I have two statements about sets which are to be verified or falsified.

  • $((a \in A) \wedge (B \subseteq A)) \Rightarrow a \in A \cup B$

  • $(A\setminus B= \emptyset) \Rightarrow A = B$

For 1: The statement is wrong if $a \in A$ and $a \notin B$.

For 2: This statement is also false. It would be correct for $(\dots) \Rightarrow A \subseteq B$.

But how do I prove this mathematically correct ?

Thank you

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For 1, the statement would have been false if it instead read $\Rightarrow a\in A\cap B$, but it doesn't read that way. It reads $a\in A\cup B$.

Remember that unions are very much like the logical "OR." $a\in A\cup B$ is the same as saying $a\in A$ or $a\in B$.

If $a\in A$ and (whatever else, doesn't matter), then in particular $a\in A$. This implies that $a\in A$ or (anything else, doesn't matter). So, statement one is in fact true.

To be more precise for a disproof for the second question, it is best to construct an explicit counterexample. As you correctly note, the hypothesis would imply $A\subseteq B$. Try to come up with a specific $A$ and a specific $B$ such that $A\subsetneq B$ (implying that $A\neq B$)

For example, $A=\emptyset$ and $B=\{1\}$