In this video on PBS Infinite Series, they attempt to prove: $$|A| < |\mathcal{P}(A)| \tag{1}$$ Where $A$ is a set (possibly infinite), $|A|$ is the cardinality of $A$ and $\mathcal{P}(A)$ is the power set of A.
They use the following argument:
Show that $|A| \le |\mathcal{P}(A)|$ (this is not relevant to my question)
Show that there is a subset $B$ from $A$ that cannot be obtained by any bijection $G:A\to \mathcal{P}(A)$, thus $|A| \lt |\mathcal{P}(A)|$
To prove 2, they present a subset $B$
$$B = \{a\in A | a\not\in G(a)\} \tag{2}$$
They then show that $$\forall a \in A, G(a) \not= B \tag{3}$$
My objection is that $(3)$ does not prove $(1)$ because $B \not\in \mathcal{P}(A)$ unless $B = \emptyset$ in which case a bijection can map any element $a \in A$ to $B$.
Since $B$ consists of all elements of $A$ for which a certain assertion holds, $B\subset A$. In other words, $B\in\mathcal P(A)$.