3.3.5.a

[ir0nt0ad]

"Let x = {{1, 2}, {1, 3}}. Then x ∈ P(A) is true. Let y = {1, 2}. Then y ∈ x and y ⊆ A are both true. Let z = {1}. Then z ∈ y and z ⊆ A are both true." It looks like z should be equal to 1, not {1}, for z ∈ y to hold.

[ir0nt0ad]

And above: "Clearly, N is a subset of its own power set, so the condition N ⊆ P(N) holds." Should it be A, not N?

[sumandroid12]

Even if so It's not "Clear" why that condition holds? Can the author please share his thoughts on why he took a long winded way of proving 3.3.4 when a short solution could have done. and how this has culminated in finding a set A other than empty set.

Let A = N∪P(N) = {∅, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, 1, 2, 3, ...}.

If A ⊆ P(A) is true. =for all x ( x ∈ A ==> x ∈ P(A)) however if x = 1 then x ∈A and x NOT ∈ P(A), P(A) contains {1} however.

[sumandroid12]

set A= {{ ϕ }, ϕ } also works. which is actually a P(A) when A = { ϕ }(which also holds this hypothesis).

In the previous exercise the author had said to prove A ⊆ P (A) → P(A) ⊆ P(P(A))

we found for A = ϕ, A ⊆ P (A) is true. thus P(A) ⊆ P(P(A)) If P(A) = x x ⊆ P (x)

and hence any set P(..P(P(A))) will hold this property as long as A=ϕ.

so base case A = ϕ P(A) = { ϕ }. P(P(A)) = {{ ϕ }, ϕ }

and so on..

[sumandroid12]

tagging @davecan author of this section.