Let S,T be sets, satisfying the property : $\forall x \in T$, $x \notin S$. Prove that $S \cap T = \varnothing$

Question

I've been thinking about this problem for roughly half an hour. I can't think of how to use subset proof structure for the empty set. Here's my answer so far:

Let $S, T$ be sets with $\forall x \in T , x \notin S$

This means that the set $S$ does not contain the element $x$. Because all of $x \in T$, $S$ and $T$ share no elements except the empty set. Hence, when $S$ and $T$ intersect, they always produce the empty set.

Am I missing the usage of definitions? And unfortunately I may not reference theorem that depict the intersection of a set with the empty set.

Answer 1

Your line of reasoning is overall correct (you understand the underlying concept), but if you were to write this as a "set inclusion" type of proof, it might go something like this:

Suppose $x \in S\cap T$. Then by definition $x \in T$ and $x \in S$, however $T$ is defined such that every $p \in T$ satisfies $p \not\in S$. Hence no element $x$ satisfies inclusion in $S\cap T$. Therefore $S\cap T = \emptyset$.

Answer 2

$\exists x.\ x \in S \cap T \implies \exists x.\ x \in S \ \wedge \ x \in T \implies \exists x \in T. \ x \in S$, which is a contradiction to your hypothesis.

Hence, $\not\exists x.\ x \in S \cap T \implies S \cap T = \emptyset.$