What are the steps of proving this?
If A\cap B' = \varnothing then $A \subseteq B$
where B' is the complement of $B$.
What are the steps of proving this?
If A\cap B' = \varnothing then $A \subseteq B$
where B' is the complement of $B$.
$x\in A \implies x\notin B^c \implies x\in B$
Let me give it a try:
Note: The conditions that A\cap B'=\varnothing and that $A\subseteq B$ are in fact equivalent.
Nana proves it directly. A proof by contradiction is supplied here.
If $A = \emptyset$, then $A \subseteq B$.
If $A \neq \emptyset$, let $x \in A$. We want to show that $x \in B$.
Suppose not, then x \in B'. We have x \in A \cap B' = \emptyset which is not possible.
Hence $A \subseteq B$.
Edit: I answered this question to test the proof approach and technique I learnt from school years ago. Would anybody point out anything that is wrong or inappropriate in the proof so that I can improve them? Thanks.