Exists Element Not in Set/Proof 2

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.

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 $S$.

$\blacksquare$

1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 2$: The Axiom of Specification