Proving that a set is empty (or is a subset to $\varnothing$)

Solution 1:

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 $x$.

Solution 2:

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.

Solution 3:

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.

Solution 4:

$A=\{x\}$ 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.