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As said in the title, this the problem I'm supposed to solve (really feeling rather stupid even asking this question, but oh well..):

Let there be sets $A$, $B$, $C$. Find all sets $X$ so that $A\cap X=B$ and $A\cup X=C$

I am very new to all of this, so I don't know how to write down what I'm thinking. In my mind I get to the point that all sets $X$ are defined here as $X \setminus (A\setminus A\cap X)$. But perhaps I am thinking the wrong way and not at all getting what I'm asked to do. Any help in this matter?

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  • $\begingroup$ do you mean $A\cup X=C$ ? $\endgroup$ – Denis Sep 16 '14 at 9:38
  • $\begingroup$ @Denis yes, perhaps that was what my prof. meant, and I might have written it down wrong, it would make more sense. $\endgroup$ – aio Sep 16 '14 at 9:41
  • $\begingroup$ You definately need $B\subset A\subset C$ if you want any solutions. $\endgroup$ – AlexR Sep 16 '14 at 9:52
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Hint: you must have $B\subseteq A\subseteq C$ and also $B\subseteq X\subseteq C$, so the problem is equivalent to find all sets $Y$ such that $$ (A\setminus B)\cap Y=\emptyset \quad\text{and}\quad Y\cup(A\setminus B)=C\setminus B. $$ The sets $X=Y\cup B$ will solve your problem, provided $B\subseteq A\subseteq C$ holds to begin with, otherwise no solution exists.

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If $B \subset A \subset C$ does not hold, there is no such $X$. Otherwise, show by double inclusion that $X = B \cup (C \setminus A)$.

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If $B$ is not a subset of $A$ then no set $X$ will exist with $A\cap X=B$, so let us assume that $B\subseteq A$ here.

If $A$ is not a subset of $C$ then no set $X$ will exist with $A\cup X=C$, so let us assume that $A\subseteq C$ here.

$X$ is a subset of $C$ with $C\backslash A\subseteq X$, $B\subseteq X$ and $(A\backslash B)\cap X=\emptyset$. That leads to $X=B\cup (C\backslash A)$

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