Find all the solutions N of the equality $\{a,b\} \cup N = \{a,b,c\} \cap N$, where $a, b$ and $c$ are some numbers I'm trying to answer the following question:
Find all the solutions N of the equality $\{a,b\} \cup N = \{a,b,c\} \cap N$ where $a, b$ and $c$ are some numbers
From what I can see, we can know that the elements $a$ and $b$ are a part of N so one solution would be $N=\{a,b\}$, although it is rather trivial. We can also have the solution $N=\{a,b,c\}$ which also seems pretty trivial. So, I'm wondering if there's a way I can find if there are more solutions to this equality since it seems like there's only two solutions. 
 A: Note that the equation implies that $\{a,b\} \cup N\subseteq N\subseteq \{a,b,c\}$ so $N$ contains $\{a,b\}$ at least and is contained $\{a,b,c\}$. This narrows down the options to the two you already found. 
A: It can be quite reduced looking at the sizes of the sets.
$| \{a, b, c\} \cap N | \leq 3$, since an intersection has at most as many elements as the set with the least elements in the intersection. This is because the two sets cannot have more elements in common than elements in one of the sets.
Due to the equality, $|\{a, b\} \cup N | = | \{a, b, c\} \cap N |  \leq 3$.
Also, since we are joining two sets, $| \{a, b\} \cup N | $ is at least the size of either of the sets. Then, $|N| \leq |\{a,b\} \cup N| \leq 3 \Longrightarrow |N| \leq 3$.
Using the same reasoning, $ 2 \leq | \{ a, b \} \cup N | = |\{a, b,c\} \cap N | \leq | N| \Longrightarrow |N| \geq 2$
This makes our problem substantially easier, since we have that $2 \leq |N| \leq 3$.
If $|N| = 2$, the intersection must contain $\{a, b\}$ because in the left-hand side it appears in a union. Knowing that $|N|=2$, we get that $N:=\{a, b\}$.
If $|N|=3$, we still get, for the same reason, that $\{a, b\}$ must be in $N$. We only have a spare element which we don't know what it is. However, it must be $c$ because the element is added in the union of the left-hand side, so it must appear in the intersection.
Then, there are only two solutions: $N:=\{a, b\}$ and $N:=\{a, b, c\}$.
