# Prove that $(A \cup B) \cap C \subseteq A \cup (B \cap C)$ [duplicate]

I'm trying to practice proof writing, and found the following text question:

For all sets A,B,C:

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

The first step I was thinking of showing is that:

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

or the subset is the sliver only where A and B intersect C.

I'm kind of stuck on where to go from here. I know the statement is true as the superset contains ALL of A, and the parts where C and B intersect, but I'm not really sure how to prove this. Any tips?

## marked as duplicate by user147263, user223391, Michael Medvinsky, Michael Burr, DavidDec 18 '15 at 1:29

You can do it directly: $$x\in(A\cup B)\cap C\Rightarrow x\in A\cup B\wedge x \in C$$Now look at the union. If $x\notin B$, it follows that $x\in A$ so $x\in A\cup(B\cup C)$. If $x\in B$, from the given fact $x\in B\wedge x\in C\Rightarrow x\in B\cap C\Rightarrow x\in A\cup (B\cup C)$ as required.

Your first step is correct. Next, notice that $A \cap C \subseteq A$. Now you can apply "${} \cup S$" to both sides of that relation, for any set $S$.

Added: By "apply '${} \cup S$' to both sides" I meant writing $(A \cap C) \cup S \subseteq A \cup S$. Taking $S = B \cap C$ gives $(A \cap C) \cup (B\cap C) \subseteq A \cup (B\cap C)$, as required. What I was really trying to communicate is that set union preserves subsets.

• Why would you add the union $S$? The second step completes the proof. – Meecolm Dec 8 '14 at 6:05
• I'm also confused at the $\cup S$ part, Couldn't I just state that since additionally $B \cap C$ is a proper subset of $A \cup (B\cap C)$ by definition of proper subsets, then close with $(A\cap C)\cup (B\cap C)$ would be a combined subset by definition of union? – Action Camu Dec 8 '14 at 6:11
• Well, I intended for the asker to replace $S$ with $(B\cap C)$. It took me personally a while to understand, back when I was learning this, that $X \subseteq Y$ implies $X \cup S \subseteq Y \cup S$ for any $S$. – Unit Dec 8 '14 at 6:14