Followup to the accepted answer of this question Direct proof of empty set being subset of every set
I understood the answer based on the nature of vacous truth, however what if we verify a statement $ \forall A \forall x, x\in \emptyset $ it holds that $x \notin A$ then this statement also becomes vacously true. Now this means $x \notin A$ is true as well.
Am I missing something, could you please explain?
On
If I've understood correctly, you are concerned about the statement $$\forall A, \forall x \in \varnothing : x \notin A.$$
Indeed, since $\forall x \in \varnothing$ is a statement about nothing, the statement is vacuously true. However, it does not mean that $\forall x : x \notin A$. In fact, the statement says pretty much nothing about $A$. To see why, we can write the statement above without syntactic sugar; it then becomes $$\forall A, \forall x : x \in \varnothing \Rightarrow x \notin A.$$
Since the premise $x \in \varnothing$ is always false, the implication is always true, and so the entire quantified statement is true. Note how the truth of this statement says nothing definite about $A$.
Let me know if I've misunderstood your question.
If you mean that every member of the empty set is not a member of $A$, then yes, you are correct. This statement is vacuously true.
There are no members to the empty set.
The point is that the definition of $X\subseteq Y$ is universal, i.e. given by $\forall x$, whereas the definition of $X\nsubseteq Y$ is existential, i.e. given by $\exists x$. So in order to prove that $\varnothing\nsubseteq A$, one needs not to verify that $\forall x\in\varnothing:x\notin A$, but rather than $\exists x\in\varnothing: x\notin A$, and the latter is indeed false.