2
$\begingroup$

Prove that $A$ and $B$ are disjoint if and only if $A\subseteq B'$

What I know:

If $S\cap T =\emptyset$ then $S$ and $T$ are said to be disjoint.

The intersection of two sets $S$ and $T$ is the set $S\cap T$, consisting of all elements that are both in $S$ and $T$, hence $S\cap T =$ {$x | x\in S$ AND $x\in T$}.

$\endgroup$
1
$\begingroup$

Maybe something along the lines of

if $A\subset B'$: $$ A\cap B' = A $$ therefore $$ A\cap B = \left(A \cap B'\right)\cap B = A \cap \left(B'\cap B\right) = A \cap \emptyset = \emptyset $$

if $B' \subset A$ then there $$ A\cap S = A = A\cap(B'\cup B) = (A \cap B') \cup (A \cap B) = B' \cup (A \cap B) $$ if $A \cap B = \emptyset$ then $A = B'$ which contradicts the first statement of $B' \subset A$

for $A = B'$ it is easy to sure that this disjoint.

$\endgroup$
0
$\begingroup$

Here is a simple proof. Let's start in the forward direction.

Suppose $A$ and $B$ are disjoint. Then $\forall a \in A$, $a \notin B \Longrightarrow a \in B'$. Thus we have that $A \subseteq B'$, because every element of $a$ must also be in $B'$.

Now to prove the reverse statement, let $A \subseteq B'$. Then $\forall a \in A$, we also have $a \in B' \Longrightarrow a \notin B$. Similarly, if $b \in B$ then $b \notin B'$, so especially we cannot have $b \in A$. Therefore $A$ and $B$ are disjoint.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.