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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 $A=\varnothing$?

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  • $\begingroup$ The empty set is a set without any element in it. Hence the contradiction works if you follow this simple definition. $\endgroup$
    – nb1
    Nov 12, 2014 at 12:52
  • $\begingroup$ What do you know about set $A$? $\endgroup$
    – Thumbnail
    Nov 12, 2014 at 12:57
  • $\begingroup$ @Thumbnail As a simple example: $\forall B. A = B \setminus B$. $\endgroup$ Nov 12, 2014 at 12:59
  • $\begingroup$ Your suggestion of proving $\forall x. x\in A \implies x \in\varnothing $ does make sense. $x\in\varnothing $ is always false, so this is the same as showing that an assumption of $x\in A $ leads to a contradiction. This answer discusses a similar situation. $\endgroup$
    – MJD
    Nov 12, 2014 at 13:15
  • $\begingroup$ @MJD Oh exactly, that's why I'm confused. But to prove that $A = \varnothing$, don't I have to prove that? $\endgroup$ Nov 12, 2014 at 13:16

4 Answers 4

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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 $x$.

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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.

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  • $\begingroup$ In the body of my question I'm describing that I'm having a hard time formally proving $\varnothing \subseteq A$. $\endgroup$ Nov 12, 2014 at 13:18
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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.

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$A=\{x\}$ 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.

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