I need help solving this question. I understand that we need to prove that the LHS is a subset of the RHS and that the RHS is a subset of the LHS, but I don't really know what to do afterwards.
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2You might want to include your attempts so that this doesn't become a "solve my homework" kind of question. – Promethèus Jul 27 '23 at 21:34
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1Does this answer your question? Prove $ (A \cup B) \cap C$ = $(A \cap C) \cup (B \cap C) $ – A. Goodier Jul 27 '23 at 21:38
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@A.Goodier Nope. Sorry, but it does it in another way rather than the way I'm trying to solve it. – CodingKook Jul 27 '23 at 22:02
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@DiegoCimadom Yeah. You're probably right. However, I'm still a beginner when it comes to proofs, so I'd like to just understand how to write the solution first. – CodingKook Jul 27 '23 at 22:03
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@CodingKook Also see the linked questions on that post. This question has been asked many times before. – A. Goodier Jul 27 '23 at 22:10
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First, take an arbitrary element $x \in (A \cup B) \cap C.$ This is the same as saying that $x \in A \cup B$ and $x \in C.$ Using the definition of union of sets, we have that either $x \in A$ or $x \in B.$ In the first case, we have that $x \in A$ and $x \in C,$ i.e., $x \in A \cap C.$ In the latter, we have that $x \in B$ and $x \in C$, i.e., $x \in B \cap C.$ In both of these cases we reach the conclusion that $x \in (A \cap C) \cup (B \cap C),$ just by definition of union of sets. This is enough to show that $$(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C).$$
On the other hand, consider an arbitrary element $x \in (A \cap C) \cup (B \cap C).$ Then, we have that either $x \in A \cap C$ or $x \in B \cap C.$ In the first case, we have $x \in C$ and $x \in A$. Note that since $x \in A$ then it follows that $x \in A \cup B.$ Therefore, it follows that $x \in (A \cup B) \cap C.$ In the latter, we have $x \in C$ and $x \in B.$ Again, note that $x \in B$ implies that $x \in A \cup B$ and therefore we have that $x \in (A \cup B) \cap C.$ With both cases in mind, we've just proved that
$$ (A \cap C) \cup (B \cap C) \subseteq (A \cup B) \cap C, $$
proving the equality of sets you're looking for.
xyz
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Would you mind explaining this if it's not a problem? - Note that since x∈A then it follows that x∈A∪B – CodingKook Jul 27 '23 at 22:06
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Sure. This conclusion follows directly from the definition of union of sets. We define $$ A \cup B = { x , \colon , x \in A \text{ or } x \in B }. $$ Therefore, if you know that $x \in A,$ the condition $x \in A \text{ or } x \in B$ is verified, which means that $x \in A \cup B.$ – xyz Jul 27 '23 at 22:26
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The first inclusion can be shown as follows:
\begin{align} x \in (A \cup B) \cap C &\rightarrow x \in A \cup B \wedge x \in C\\ &\rightarrow (x \in A \lor x \in B) \wedge x \in C\\ &\rightarrow (x \in A \wedge x \in C) \lor (x \in B \wedge x \in C)\\ &\rightarrow (x \in A \cap C) \lor (x \in B \cap C)\\ &\rightarrow x \in (A \cap C) \cup (B \cap C). \end{align} Therefore, $(A \cup B) \cap C \subseteq (A \cap C) \cup (B \cap C)$. The other inclusion is proved similarly.
Promethèus
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