Confusion in understanding why empty set is a subset of every set statement 1 : If $x \in \emptyset$ then $x \in A$.
statement 2 : If $x \in \emptyset$ then $x \notin A$.
I know that both statements are true since the hypothesis is false.
But first statement says that $\emptyset \subset A$ while second statement says that $\emptyset$ is not a subset of $A$.
My question is why we prefer $\emptyset \subset A$ over the other implication that $\emptyset$ is not a subset of $A$? Thanks.
 A: The statement

(For all $x$,) if $x\in B$ then $x\notin A$.

does not say that $B$ is not a subset of $A$. In order to say that $B$ is not a subset of A, you would need to say

There is an $x\in B$ such that $x\notin A$.

This is vacuously false then $B$ is $\varnothing$.
A: Second statement shows that $$\emptyset \subseteq A^c$$ which is also true. We are not preferring one over other. Both are true.
Edit 1: The second statement does not imply that $\emptyset$ is not a subset of $A$. For that you would need that there exists $x \in \emptyset$ such that $x \in A$. This is indeed false.
Edit 2: What second statement implies is $\emptyset \cap A=\emptyset.$
A: In elementary set theory, each set can be defined as $A=\{x\mid P(x)\}$ where $P(x)$ is a predicate with free variable $x$, and each subset as $B=\{x\mid x\in A\wedge Q(x)\}$ with predicate $Q$.
Define $\emptyset_A=\{x\mid x\in A\wedge x\not\in A\}$. This is a subset of $A$, the empty subset of $A$. For any other set $B$, define similarly $\emptyset_B=\{x\mid x\in B\wedge x\not\in B\}$ as the empty subset of $B$.
But $\emptyset_A=\emptyset_B$, since $x\in\emptyset_A\Leftrightarrow x\in A\wedge x\not\in A\Leftrightarrow x\in B\wedge x\not\in B\Leftrightarrow x\in\emptyset_B$. In the 2nd equivalence, a false assertion is replaced by another false assertion.
Hence, the empty subset of a set $A$ is independent of the set $A$ itself and so we can safely speak of the empty set by putting $\emptyset=\emptyset_A$.
