can someone help me with this problem.
Notation: $A$' is the complement of A
Prove that $A$ $\cap$ $B$' $\subseteq$ ($A$ $\cap$ $B$)'
My work:
$A$ $\cap$ $B$' = {x $|$ x $\in$ A, x $\notin$ $B$}
$A$ $\cap$ $B$ = {x $|$ x $\in$ A, x $\in$ $B$}
($A$ $\cap$ $B$)' = {x $|$ x $\notin$ A, x $\notin$ B} = {x $|$ x $\notin$ ($A$ $\cap$ $B$)}
I am stuck here. Please help if you can. Thank you!
@DeepSea is correct. I just want to touch on something in your solution which may give you insight with future problems you try to solve. There is a problem in your understanding with the last line - $(A \bigcap B)'$ means that $x$ is not in the compliment of the intersection. So $x$ could very well be in $A$ or $B$ - just not both $A$ and $B$.
If $x \in A \cap B’ \implies x \in A$ and $x \in B’$. If $x \in A\cap B$ then $x \in B$ since $A \cap B \subseteq B$, but $x \notin B$ so $x \in (A \cap B)’$. Thus: $A \cap B’ \subseteq (A \cap B)’$.
A basic fact about the $'$ operator is that it is decreasing - that is, if $X \subseteq Y$ then $Y' \subseteq X'$.
A basic fact about the $\cap$ operator is that $X \cap Y \subseteq Y$.
We note by the $\cap$ basic fact that we have $A \cap B \subseteq B$. Therefore, by the basic fact about $'$, we have $B' \subseteq (A \cap B)'$. Using the $\cap$ basic fact again, we have $A \cap B' \subseteq B'$. Therefore, we have $A \cap B' \subseteq B' \subseteq (A \cap B)'$. Then $A \cap B' \subseteq (A \cap B)'$.
