Elementary set theory Exercise Let $A$ ,$ B$  and $ C $ be three parts of a set E such that: $ B ⊂ A ⊂ C $
Determine the set $X ⊂ E $ knowing that: 
$A \cap X = B$ and  $A \cup X = C $
I tried to prove that $X = (C ∩ \overline A) ∪ B$:
$A ∪ X= C ⇒ (A ∪ X) ∩ \overline A = C ∩  \overline A ⇒ X ∩  \overline A = C ∩  \overline A $
$⇒ (C ∩  \overline A) ∪ B= (X ∩  \overline A) ∪ B = (B ∪ X) ∩ ( \overline A ∪ B)= X ∩ ( \overline A ∪ B) $(since $B  ⊂ X$)
Let's  prove that$ X ⊂ ( \overline A∪ B)$:
$A ∩ X = B ⇒ (A ∩ X) ∪  \overline A = B ∪  \overline A ⇒ X ∪  \overline A = B ∪  \overline A ⇒  X ⊂ ( \overline A∪ B)$
Therefore$ (C ∩  \overline A) ∪ B= X$
My question is:
First of all, have I committed any mistakes?
Second of all, Is there any solution we can write  without using Venn diagram?( Because I've used it beforehand to know that $X = (C ∩ Ac) ∪ B) $) if so, any input would be appreciated.
 A: There’s nothing wrong with using a Venn diagram to see what’s going on. However, you can avoid it here if you notice that


*

*$A\cap X=B$ implies that $B\subseteq X$ and $X\cap(A\setminus B)=\varnothing$, and

*$A\cup X=C$ implies that $X\subseteq C$ and $X\supseteq C\setminus A$.


Since $B\subseteq A\subseteq C$, $C$ can be partitioned into $B$, $A\setminus B$, and $C\setminus A$. $X$ is a subset of $C$ that contains the first and last of these pieces and is disjoint from the second, so it must be exactly the union of the first and last pieces: $X=B\cup(C\setminus A)$.
Your proof is also fine. You could also use a point-chasing argument, which is often easier. Suppose that $x\in X$; then $x\in C$, since $A\cup X=C$ and therefore $X\subseteq C$. Now either $x\in A$, or $x\notin A$. If $x\in A$, then $x\in B$, since $A\cap X=B$. If $x\notin A$, then $x\in C\setminus A$. In either case $x\in B\cup(C\setminus A)$, so $X\subseteq B\cup(C\setminus A)$.
Now suppose that $x\in B\cup(C\setminus A)$, so that $x\in B$ or $x\in C\setminus A$. $A\cap X=B$, so $B\subseteq X$, and therefore $x\in B$ implies that $x\in X$. If, on the other hand, $x\in C\setminus A$, then $$x\in(A\cup X)\setminus A=X\setminus A\subseteq X\;,$$ so again $x\in X$, and we’ve shown that $B\cup(C\setminus A)\subseteq X$ and hence that $X=B\cup(C\setminus A)$.
