Proving that a set is empty (or is a subset to $\varnothing$) I want to prove that set $A$ is empty ($A = \varnothing$).
Intuitively, I understand that a way to do it would to be to falsely assume $ x \in A$ and show a contradiction (because if it's empty, $x \notin A$).
But I don't understand how that works mathematically.
Mathematically, I'm tripping because I need to prove is $A \subseteq \varnothing$, which would mean that $\forall x. \text{ if } x \in A \text{ then } x \in \varnothing$ (I'm aware that doesn't makes sense, but still required, which is why I'm confused).
According to this question, the contradiction really is enough, but why? how does that help you prove $A \subseteq \varnothing$, for you to be ultimately able to say $A=\varnothing$?
 A: Because $x \in \varnothing$ is identically false, the only way $x \in A \implies x \in \varnothing$ can be true is if $x \in A$ is false.
Thus, the only way $\forall x: x \in A \implies x \in \varnothing$ can be true is if $x \in A$ is false for all $x$.
A: If you prove $\boldsymbol{A}\subseteq \boldsymbol{B}$ and $\boldsymbol{B}\subseteq \boldsymbol{A}$ then you know $\boldsymbol{A} = \boldsymbol{B}$.
Hint: 
The $\varnothing \subseteq \boldsymbol{A}$ by definition of being the empty set.
A: This is essentially a proof by contraction. In a proof by contradiction, you assume some assertion P is true, and then deduce a contradiction from it. You may then conclude P is false, as if it were true, a statement known to be false would be true.
To prove the set A is empty, begin by assuming A is non-empty. Using existential-instantiation, you may then define x to be an element of A (since you've assumed at least one exists). If you can then derive a contradiction from the assumption x is an element of A, the original supposition that A is non-empty must be wrong, and you may conclude A is empty.
"Mathematically, I'm tripping because I need to prove is A⊆∅, which would mean that ∀x. if x∈A then x∈∅"
Another way to look at it is that since anything follows from a contradiction, if you can prove a contradiction follows from the assumption x∈A, then you can prove anything including x∈∅ follows from the assumption x∈A.
A: $A=\{x\}$
but $A$ is a subset of  and equal to empty set:
empty set$=\{\}$
therefore, $A=$ empty set$=\{\}$ or $\{x\}$
but empty set is also a subset of $A$
therefore, probability of $A$ tending to null$=1$;also probability of $A$ tending to $\{x\}=1$
therefore, $A$ can be $= \{x\}$
$A$ can be$=\{\}$
therefore $A=\{\}$ because the possibility of $A$ tending to $\{\}$ is higher if and only if limits is involved.
