Prove that for all sets $A$ and $B$, $( A \cup B ) - B = A$ implies $A \cap B =\varnothing$.

Question

In the following proof we harness properties such as: For all set $A$,

(i) $A = A - \varnothing$

(ii) $A \cap\varnothing=\varnothing$

Proof:

Let $A$ and $B$ be arbitrary sets and suppose $( A \cup B ) - B = A$.

Given (i) then $$( A \cup B ) - B = A - \varnothing.$$

So then, $( A \cup B ) = A$ and $B = \varnothing$.

That being so, $$A \cap B = ( A \cup B ) \cap \varnothing = \varnothing$$

If we apply (ii) to the foregoing equivalence relation. Therefore $$A \cap B = \varnothing$$

Is this proof right? if it is I would like to see alternative ones.

Answer 1

Suppose that $x \in A \cap B$ existed. Then $x \in A$ but also $x \in A \cup B$ and $x \in B$ so that $x \notin (A \cup B) - B$. This contradicts the starting assumption that $A=(A\cup B)-B$, so $A \cap B = \emptyset$.

Answer 2

So then, ( A ∪ B ) = A and B = ∅.

No, that is not valid.   For instance: $(\{1,2,5\}\cup \{3,4\})\smallsetminus\{3,4\}=\{1,2,5\}\smallsetminus \emptyset$ does not entail that $\{3,4\}=\emptyset$

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Hint: Show by definition of set difference and distribution that : $(A\cup B)\smallsetminus B = (A\smallsetminus B)\cup(B\smallsetminus B)$. Thus since the LHS equals $A$, therefore....

Answer 3

By definition of $X=C-B$, no element of $X$ lies in $B$. Hence, if $A=(A\cup B)-B$, no element of $A$ lies in $B$, and of course, conversely no element of $B$ lies in $A$, which means $A\cap B=\varnothing$.

Answer 4

Suppose $(A\cup B) \setminus B = A$.

Let $x \in A\cap B$.

Then $x \in A$ and $x\in B$. Then $x \in B$ so $x\not \in M\setminus B$ for any set $M$ so $x\not \in (A\cup B) \setminus B$. So $x \not \in A$. But that's a contradiction.

So if $(A\cup B)\setminus B =A$ then $x \in A\cap B$ is impossible. So $A\cap B=\emptyset$.

Not this in no way means $B$ is empty; just $A$ and $B$ are dijoint.

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Alternatively. If $(A\cup B)\setminus B = A$ then if $x \in A$ then $x\in (A\cup B) \setminus B$ so $x \not \in B$. So $x$ in both $A$ and $B$ is impossible so $A\cap B=\emptyset.

Or....

If $(A\cup B)\setminus B = A$ and if $x \in B$ then $x\not \in (A\cup B)\setminus B = A$ so $x$ in both $B$ and $A$ is impossible.

....

An alternative way would be to prove that $(M\cup N)\setminus N = M\setminus N$ always. I'll leave that proof to you.

That means $(A\cup B) \setminus B= A\setminus B= A$. So if $x\in B$ the $x$ is removed for $A$ to get $A\setminus B$ so $x\not \in A$.

Etc.