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I have a question in regard to some questions I am working on in beginner set theory.

I will give an example to illustrate by question better,

For example, say we were wanting to prove

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

I know that to do this I must show that they are both subsets of each other.

But my question is mostly about the operations that are allowed,

say we are first wanting to show

$$A \cup(B \cap C) \subset (A \cup B) \cap (A \cup C)$$

Now, intuitively and from previous basic proofs I know that

$B \cap C \subset B$

and $B \cap C \subset C$

Now, here is my real question, can we treat these type of problems and proofs and unions/intersections as something like a linear operator, that is , would it always be valid to say from

$$B \cap C \subset B$$ that $$A \cup (B \cap C) \subset A \cup B$$

Ie, that we just added in a A union on both sides, almost like in a basic equation with additions and subtractions. But I am not sure if this works like this? Can anyone shed some info on that?

Thank you

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Yes, you can prove that given $A,B,C$ sets, we have: $$B \subseteq C \implies \begin{cases} A \cup B \subseteq A \cup C \\ A \cap B \subseteq A \cap C\end{cases}$$and use it as a theorem.

Suppose that $B \subseteq C$. If $x \in A \cup B$, we have two cases: $x \in A$, and so $x \in A \cup C$, or $x \in B$, and hence $x \in C$ (by hypothesis), and we get $x \in A \cup C$ in the same way.

Suppose that $B \subseteq C$. If $x \in A \cap B$, we have $x \in A$ and $x \in B$. By hypothesis, we have $x \in A$ and $x \in C$, hence $x \in A \cap C$.


Let $X$ be the universe. Fix $A \in \wp(X)$. We can define $f_A,g_A: \wp(X) \to \wp(X)$ by: $$f_A(B) = A \cup B \quad\text{and}\quad g_A(B) = A \cap B.$$If you order $\wp(X)$ by inclusion, we can say that $f_A$ and $g_A$ are increasing maps. Above we've shown that $B \subseteq C \implies f_A(B) \subseteq f_A(C)$ and $g_A(B) \subseteq g_A(C)$.

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  • $\begingroup$ Right, thanks I do understand that, my question is more about if we can always treat union/intersections as "operators", I don't know the right term though $\endgroup$ – Quality Jun 14 '15 at 22:51
  • $\begingroup$ I added a bit on that in the answer. $\endgroup$ – Ivo Terek Jun 14 '15 at 22:53

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