# Is following proof of $\exists X \forall Y \exists Z .(X\subseteq Y \cap Z)$ valid?

I have to make or deny proof for : $$\exists X \forall Y \exists Z .(X\subseteq Y \cap Z)$$

Is the proof by saying that $$\emptyset \subseteq All$$ and so $$X = \emptyset$$
enough?

• Could you state in english what the statement you are trying to prove is? It sounds like you are trying to prove: There exists a set $X$ so that for every set $Y$ there exists a set $Z$ so that $X\subseteq Y\cap Z$ which is... as you argue ... trivial. But a bizarre statement (who cares about the $Z$?) – fleablood Nov 29 '19 at 17:29
• yeah.. It sound very trivial.. In fact, it's not mine, I found it in one exercise book online and it just seems so easy, that I'm confused a little bit .. There is nothing more than this the exercise gave me.. – Patrik Bašo Nov 29 '19 at 17:30
• Well, Sure $X=\emptyset$ will be a subset of $Y\cap Z$ for any $Y$ and any $Z$. And it's the only such set the statement is true. If $X$ is non empty then there is an $x \in X$ an there can exists sets $Y$ so that $x \not \in Y$ and so $X \not \subset Y$ and nor can $X\subset W\subset Y$ for any subset, $W$, of $Y$. So there can not exist any set $Z$ so that $X\subset Y\cap Z$.... I guess it's a valid excercise in concept and definitions .... but basic. You seem to be beyond the level of such definition quiz cards. – fleablood Nov 29 '19 at 17:37
• Yeah. exactly. I must find some more difficult ones – Patrik Bašo Nov 29 '19 at 17:39

## 1 Answer

Yes it is, there is no trick question.