For any sets $A, B, C$ within a universal $U$ set, prove that $A\cup B \subseteq C$ iff $(A \cup C)\cap (B \cup C) = U$

Question

For any sets $A, B, C$ within a universal $U$ set, prove that $A\cup B \subseteq C$ iff $(A \cup C)\cap (B \cup C) = U$

Confused on how to do this, any help would be great.

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Correction: Accidentally had it say = C when it should have said = U. Sorry!

Answer 1

The statement is false; @Daniel Schepler provided a counterexample in the comments. The statement

$$(A \cap B = C) \implies ((A \cup C) \cap (B \cup C) = C)$$

is true; we can show it by assuming that $A \cap B = C$. Let $x \in (A \cup C) \cap (B \cup C)$. Then $x \in A \cup C$ and $x \in B \cup C$. The first requirement has that $x \in A$ or $x \in C$; the second requirement has that $x \in B$ or $x \in C$. If $x \in C$, then the inclusion $(A \cup C) \cap (B \cup C) \subseteq C$ follows immediately. On the other hand, if $x \in A$ and $x \in B$, then $x \in A \cap B \subseteq C \implies x \in C$, as desired. Now let $x \in C$. It is trivial to see that this implies both that $x \in A \cup C$ and that $x \in B \cup C$, so that $x \in (A \cup C) \cap (B \cup C)$, showing the reverse inclusion. Therefore, $(A \cup C) \cap (B \cup C) = C$.

The converse

$$((A \cup C) \cap (B \cup C) = C) \implies (A \cap B = C)$$

is not even true. For take $A = \{ 1, 2 \}$, $B = \{ 2, 3 \}$, and $C = \{ 2, 4 \}$. Here, $A \cap B = \{ 2 \} \ne C$. However, $(A \cup C) \cap (B \cup C) = \{ 1, 2, 4 \} \cap \{ 2, 3, 4 \} = \{ 2, 4 \} = C$.

Answer 2

I don't think it is true. Take $A=B=\{x\}$, $C=\{x, y\}$, $U=\{x,y,z\}$.

Then $(A \cup C) \cap (B \cup C) = C \subsetneq U$.