Exists Element Not in Set/Proof 2
Jump to navigation
Jump to search
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
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 $S$.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 2$: The Axiom of Specification