Let $A, B$, and $C$ be sets. If $A ⊆ C$ and $B ⊆ C$, then $A ∩ B ⊆ A ∩ C$

Question

I understand the definitions of the set operations, but I'm stuck actually proving the statement.

What I have:

If $A ⊆ C$ , and $B ⊆ C$, every element from both $A$ and $B$ is in $C$. What I think is tripping me up is the fact that $A$ and $B$ don't necessarily have to have any common elements with each other.

For Instance, if $A = \{1,2\}$, $B = \{3,4\}$, and $C = \{1,2,3,4\}$, both $A$ and $B$ are subsets of $C$, yet $A ∩ B = \varnothing$.

However, even in this instance, wouldn't $A ∩ B ⊆ A ∩ C$ still be true, since the empty set is a subset of every set? This is where I'm confused: is this evidence that the statement is true and the proof is complete? Or is it not valid?

Answer 1

Correct: If $A\subseteq C$ and $B\subseteq C$ then any elements in both $A$ and $B$ are in $C$.   That is $A\cap B\subseteq C$.

Also $A\cap B\subseteq A$, so then...

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Addendum, the empty set is a subset of any set, so the above works when $A\cap B=\emptyset$, since $\emptyset \subseteq C$ and $\emptyset\subseteq A$.

Answer 2

The proof will still be okay, basically for the reason you described.

Think about it this way: when you're proving $S \subseteq T$, you're proving (for all $x$) a conditional $x \in S \implies x \in T$. If any particular $x \notin S$ -- or even if every single $x \notin S$, i.e., the case that $S = \varnothing$ -- then the conditional is vacuously true.

Answer 3

Actually, the assumption that $A\subseteq C$ is not necessary. As long as $B \subseteq C$ then for any set $A$, we have $A\cap B\subseteq A\cap C$. To prove this, we need the following cases:

Case 1. Suppose that $A\cap B=\varnothing$. Then we are done.

Case 2. Suppose that $A\cap B\neq\varnothing$. Let $x\in A\cap B$. Then $x\in A$ and $x\in B$. Since $B\subseteq C$ and $x\in B$, it follows that $x\in C$. This shows that $x\in A\cap C$ and we are done.

Answer 4

That's a little misleading. It is true that $A\cap B\subset A\cap C$ but its also true tht$A\cap B \subset B\cap C$.

Whenever $X \subset Y$ then $X\cap Y = X$. And for any two sets $X, Z$ we always have $X \cap Z \subset X$ and $X \cap Z \subset Z$.

So as $A$ and $B$ are subsets of $C$ we know:

$A\cap C = A$ and $B\cap C = B$ and $A\cap B \subset A = A\cap C$ and $A\cap B \subset B = B \cap C$.