Proving $A \cap C = B \cap C$, but $ A \neq B$

Question

Let $A,B,C$ be sets. Identify a condition such that $A \cap C = B \cap C$ together with your condition implies $A=B$. Prove this implication. Show that your condition is necessary by finding an example where $A \cap C = B \cap C$, but $ A \neq B$

Edit: I've read the wrong proposition/definition. UGH! The question was probably about having sets being equal, not empty.

Now, suppose that $ A \neq B$... that would mean that $A$ isn't equal to $B$. So they are different sets, but $A \cap C = B \cap C$ are equal sets.

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I'm lost on this. I need to find a condition, but where do I even start? These are my thoughts about the question so far.

We need to find a condition for $A \cap C = B \cap C$.

Proposition 3.1.12 states that if $A$ and $B$ are both empty sets, then $A =B$.

So $A \cap C$ and $B \cap C$ are both empty sets which is why $A \cap C = B \cap C$.

It seems that there are empty sets everywhere because proposition 3.1.12 claims that if $A$ and $B$ are empty set, then $A =B$. It's like there aren't any elements at all. There's nothing.

We need to find a condition that demonstrates that $A \cap C = B \cap C$ which are empty sets, but $A \neq B$ means that there are elements in $A$ and $B$.

How is this even possible?

$A \cap C$ by definition 3.2.1 is $[x: x: \in A \land x \in C]$

x belongs in A and x belongs in C.

$B \cap C$ by definition 3.2.1 is $[x: x: \in B \land x \in C]$

x belongs in B and x belongs in C

The only way I could think of is a contradiction to this... but that would mean that $A = B$ ... there are no elements in A and B, but $A \cap C \neq B \cap C$ means that there are elements. There may not be elements in A and B, but there are elements in C.

Answer 1

HINT: What happens if both $A=A\cap C$ and $B=B\cap C$? (Note that this generalizes the case of both $A$ and $B$ being empty.)

Answer 2

If $A \not\subset C$, then let $x \in A \setminus C$, and $B = A\setminus \{x\}$. Then you can see that $A\ne B$ and $A \cap C = B \cap C$. Hence you must have $A \subset C$. By symmetry, we must have $B \subset C$. Combining shows that you need $(A \cup B) \subset C$.

Now suppose $(A \cup B) \subset C$. Then $A \cap C = A$ and $B \cap C = B$.

Answer 3

The condition for having

$A \cap C = B \cap C$ but at the same time $A \neq B$

impose you that

$A \cap C = \emptyset = B \cap C$

This does not imply that $A$ and $B$ are both empty.

Let $\quad C = \{ x \in \mathbb{N} : Even(x) \}$

and let $\quad A = \{ 1,3,5,7 \} \quad$ and let $\quad B = \{ 9,11,13,15 \}$.

Because both $A$ and $B$ do not contain even numbers, we have that :

$A \cap C = \emptyset = B \cap C$

but $A \neq B$.