Those are the text used in the video
For any finite sets x, we have that |P(x)| > |x|. In particular |{{}}| = |P({})| > |{}|. This is, in terms of Neumann ordinals, 1 > 0. When we say "function" in this text, we always mean function on a non-empty domain.
Consider a theory of sets and a notion of function so that, for each function f, dom(f) is a set. (E.g. when functions are modeled as special relations and relations are sets of pairs in the usual way.)
Definition of "injective":
injective(f) ≡ ∀(x ∈ dom(f)). ∀(y ∈ dom(f)). f(x) = f(y) => x = y
Onto will be defined and used later in this text.
Consider now the property on functions whos "values are subsets of its domain" V(f) = ∀(x ∈ dom(f)). f(x) ⊆ dom(f)
If the power set P_X exists as a set, we might also say that the values are in the power set. Indeed, then for any set X, consider F: X -> P_X F(x) = {x}
Proposition 1. ∀X. ∃f. V(f) and injective(f)
For a concrete case, G: N -> P_N G(n) := {n} # always a 1 element set. Also, for the same reason, clearly not onto.
E.g. G(3) = {3} G(10) = {10} But it also works if the codomain is smaller. G': N -> {x | ∃(k ∈ N). x={k} } G'(n) := {n}
Or H: Q -> P_Q H(q) := {q - 1/7, 2 * q} # always a 1 or 2 element set E.g. H(3) = {3 - 1/7, 6} H(1/7) = {0, 2 / 7} H(-1/7) = {-2/7} H(0) := {-1/7, 0}
Or J: N -> P_N J(n) := {k ∈ N | k > 2 * n } E.g. J(10) = {21, 22, 23, ...} J(1) = {3, 4, 5 ...} J(0) = {1, 2, 3, ...} # all members always bigger than 0
Consider for any F with V(f) and the following set, if it exists (very weak assumption) D_F ≡ {x ∈ dom(F) | x ∉ F(x)}
E.g. D_G ≡ {n ∈ N | n ∉ {n}} = {} # is empty D_H ≡ {q ∈ Q | q ∉ {q - 1/7, 9001 * q}} = Q \ {0} # is infinite D_J ≡ {n ∈ N | n ∉ {2n+1, 2n+2,...}} = N # holds the number 0
Note that in those cases at least, the set D_f are all different from the values of F!
Consider a function I : {3, 5, 9} -> {{5}, D_I} and let's reason about I by considering D_I D_I = {x ∈ {3, 5, 9} | x ∉ I(x)}
For a contradiction, assume I(3) = D_I Then in the comprehension above, 3 ∉ D_I <=> 3 ∈ D_I. As this goes for all n, we find I(n)={5} for all n. Hence D_I = {x ∈ {3, 5, 9} | x ∉ {5}} = {3, 7} So I : {3, 5, 9} -> {{5}, {3, 7}} but D_I is also never taken by D_I as a value.
Proposition 2: ∄f. V(f) and [ ∃(z ∈ dom(F)). F(z) = D_F ]
For a contradiction, assume ∃f. V(f) and [ ∃(z ∈ dom(F)). F(z) = D_F ] ∃f. V(f) and [ ∃(z ∈ dom(F)). F(z) = {x ∈ dom(F) | x ∉ F(x)} ] ∃f. V(f) and [ ∃z. z ∈ dom(F) and [ ∀x. x ∈ F(z) <=> (x ∈ dom(F) and x ∉ F(x)) ] ]
Take said F and said c ∈ dom(F) and consider the claim ∀x. x ∈ F(c) <=> (x ∈ dom(F) and x ∉ F(x)) Now for a particular, x=c, this then claims c ∈ F(c) <=> (c ∈ dom(F) and c ∉ F(c)) which reduces to c ∈ F(c) <=> c ∉ F(c) A contradiction. QED.
Proposition 3. ∄f. onto(f, P_dom_f)
Let onto(f, Y) ≡ ∀(x ∈ dom(f)). f(x) ∈ Y and ∀(y ∈ Y). ∃(z ∈ dom(f)). f(z) = y
For a contradiction, assume there's a surjection into the power set. I.e., with Y = P_dom_f, ∃f. onto(f, P_dom_f)
Take said F. The first half translates to V(F) and all our above considerations apply.
The second part says ∀(y ∈ P_dom_F). ∃(z ∈ dom(F)). F(z)=y Now for a particular, y=D_F, this then claims ∃(z ∈ dom(F)). F(z)=D_F
Together get get ∃f. V(f) and [ ∃(z ∈ dom(F)). F(z)=D_F ] A contradiction. QED.
We have shown existence of injection and non-existence of surjection, so classically |P_N| > |N|.