Exists Element Not in Set

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Theorem

Let $S$ be a set.


Then:

$\exists x: x \notin S$


That is, for any set, there exists some element which is not in that set.


Proof 1

Consider the power set $\powerset S$ of $S$.

Aiming for a contradiction, suppose $\forall x \in \powerset S: x \in S$.

Then the identity mapping $I_S: S \to \powerset S$ would be a surjection.

But from Cantor's Theorem, there is no surjection $f: S \to \powerset S$.

From this contradiction it follows that:

$\exists x \in \powerset S: x \notin S$

Hence the result.

$\blacksquare$


Proof 2

By Axiom of Specification, we can construct the set:

$T = \set {x \in S: x \notin x}$

Then for all $y$:

$(*) \quad y \in T$ if and only if $\paren {y \in S \land y \notin y}$.


Aiming for a contradiction, suppose $T \in S$.

By Law of Excluded Middle, either $T \in T$ or $T \notin T$.


Suppose $T \in T$.

By $(*)$, $T \in S \land T \notin T$.

By Rule of Simplification we have $T \notin T$, which is a contradiction.


Now suppose $T \notin T$.

By $(*)$ again, we have $\neg \paren {T \in S \land T \notin T}$.

By Modus Ponendo Tollens, $\neg \paren {T \in S}$, which is also a contradiction.


Hence we must have $T \notin S$.

And thus there exists something (namely $T$) that does not belong to $S$.

$\blacksquare$