Prove $D=${$x\in B:x\cap y=x$ for any $y\in A$} is the empty set $$A=\{\emptyset,\Bbb Z,\Bbb Q\},\\ B=\{\{\emptyset\},\Bbb Z,\Bbb Q\},\\ C=\{x\in B:x\cap y=x\text{ for some }y\in A\},\\ D=\{x\in B:x\cap y=x\text{ for any }y\in A\}.$$
I have to prove whether $D$ is the empty set. But I'd like to ask what's the difference between 'some' and 'any'? What is the difference when I want to proof whether set $C$, $D$ is empty or not?
Am I right to say that there are exactly $2$ elements in $C$, which are $\Bbb Z,\Bbb Q$?
 A: '$x \cap y = x$ for some $y \in A$' means 'there is an $y \in A$ such that $x \cap y = x$'. That is, $x \cap \emptyset = x$ or $x \cap {\mathbb Z} = x$ or $x \cap {\mathbb Q} = x$.
'$x \cap y = x$ for any $y \in A$' means 'for all $y \in A$ it holds that $x \cap y = x$'. That is, $x \cap \emptyset = x$ and $x \cap {\mathbb Z} = x$ and $x \cap {\mathbb Q} = x$.
A: The difference between "some" and "any" is this:

There exists a person $x$ such that $x$ is shorter than me

is a true statement, but

For any person $x$, $x$ is shorter than me

is not a true statement (I am not the tallest person in the world).

In other words, the difference is the same as the difference between the quantifiers $\forall$ and $\exists$.


Am I right to say that there are exactly $2$ elements in $C$, which are $\Bbb Z,\Bbb Q$?

Well, let's see.
Is $\mathbb  Z\in C$?
Well, we know that $x\in C\iff x\cap y=x\text{ for some }y\in A$, right? OK, that means $\mathbb Z\in C \iff \mathbb Z\cap y=\mathbb Z\text{ for some }y\in A$.
Now, does there exist some $y\in A$ such that $\mathbb Z\cap y=\mathbb Z$? Yes, there does! We can take $y=\mathbb Q$ and get $\mathbb Z\cap \mathbb Q = \mathbb Z$, which proves that $\mathbb Z\in C$.
Is $\mathbb Q\in C$?
Well, let's see. As above, we have $\mathbb Q\in C$ if and only if there exists some $y\in A$ such that $\mathbb Q\cap y = \mathbb Q$. Again, we can take $y=\mathbb Q$, and get $\mathbb Q\cap y=\mathbb Q\cap\mathbb Q=\mathbb Q$, proving that $\mathbb Q\in C$.
So, in conclusion, both $\mathbb Q$ and $\mathbb Z$ are in $C$!
Is $\{\emptyset\}\in C$?
Here, the answer shall be no. Whatever $y\in A$ we take, we will get $\{\emptyset\}\cap y=\emptyset\neq\{\emptyset\}$, so the condition for $\{\emptyset\}$ to be an element of $C$ is not fulfilled.
