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I have a task I need to solve for my set theory course. I need to prove the following formula.

$\forall A \forall a \forall b ((a \notin A \land b \notin P(A) \land |A| = |A \cup \{a\}|) \Rightarrow |P(A)| = |P(A) \cup \{b\}|)$

What I am thinking is since $|A| = |A \cup \{a\}|$ is true then A should be infinite or something else that I am missing. I would appreciate some pointers that would lead to a solution. Thank you.

Edit 1: Forgot to mention this should be solved in (ZF), so we dont have AC.

Edit 2: So having read some other posts and comments on this, let me know if I am on the right track for the problem:

This formula mainly tells us that if $A$ is infinite then its power set $P(A)$ is infinite. So we have to prove this and I was thinking of applying Cantor's theorem for this. Another thing before that is we have to prove that $\forall A \forall a(a \notin A \land |A| = |A \cup \{a\}| \Rightarrow A$ is infinite). Which seems straight forward but I guess it is not. My attempt at it is as follows:

If $|A| = |A \cup \{a\}|$ then there is a bijection between $A$ and $A \cup \{a\}$. Assume $A$ is finite, hence $(\exists n \in \Bbb N)(|A|=n)$. But $a \notin A$ so $A \cup \{a\}$ would be finite too and would have a greater cardinality then $A$, which is a contradiction.

This seems really wrong and what worries me is that I am not using the bijection I mentioned to reach the contradiction. Some clarification would be of great help and just reminding this should all be without the axiom of choice.

Ivelin Dinev
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1 Answers1

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Since A is infinite, so is P(A).
Thus |P(A)| = |P(A) $\cup$ {b}|.

Exercise: show if A is infinite, then there
is a bijection between A and A $\cup$ {a}.

To create a proof without the axiom of choice use
A is infinite iff exists bijection f:N -> A.

William Elliot
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  • I took a look at link and solutions that are both using the axiom of choice and I cannot use it for this problem. Can a simple proof be considering the example of the set of the natural numbers and the same set without zero. The latter is a proper subset of the former and $n \rightarrow n + 1$ is a bijection between them. – Ivelin Dinev Feb 12 '20 at 01:24
  • No, one example is not sufficient, See edit. – William Elliot Feb 12 '20 at 04:15
  • We do not have the axiom of choice in ZF – Ivelin Dinev Feb 12 '20 at 09:02
  • On top of that should we not prove the other way around, since we are given that $|A| = |A \cup {a}|$ and we need to prove that A is infinite. Also I edited my post according to what I think. – Ivelin Dinev Feb 12 '20 at 15:49
  • @IvelinDinev At the bottom I gave you a hint how prove it without AxC. To show A is infinite, show for all finite A, A is not equinumerous to A with an additional element. – William Elliot Feb 12 '20 at 19:30