# Dependence of the equation $a+1=a$ for infinite cardinals $a$ on the axiom of choice

let $$A$$ be a set such that for all $$n \in$$ N $$A ≉ N_n$$ where $$N_n = \{ 0 ,1 ,2 ...... n-1\}$$

and $$a$$ be the Cardinality of $$A$$ meaning ($$|A| = a$$)

is it possible to prove that $$a+1=a$$ without using Axiom of choice ?

As a student who has just completed a basic course in set theory, I find it interesting that without the Axiom of Choice, a simple assertion like $$a=a+1$$ cannot be proven..

• $a + 1 = a$ is usually called "$a$ is Dedekind-infinite" in choiceless contexts, whereas "$A$ is not in bijection with any finite set" is called "$a$ is infinite". You need some Choice to show "infinite implies Dedekind-infinite" - see Wikipedia and this and this Commented May 6 at 9:56
• @Peter That's not true. See DanielWainfleet answer. Commented May 7 at 1:21
• @Peter The issue is your claim "Adding an element does then not change the cardinality." In the absence of (a weak form of) the axiom of choice, adding a single element to an infinite set can change its cardinality. Commented May 8 at 18:59
• I would not be surprised if this question were a duplicate (although I don't offhand see a dupe target), but it certainly doesn't lack context. I've voted to reopen. Commented May 8 at 19:00
• I see that this has attracted another vote to close as "Missing context." What context, exactly, is missing here? Commented May 8 at 20:16

Without AC, one must be cautious about "finite" and "infinite". A set $$T$$ is Tarski-finite iff each non-empty family of subsets of $$T$$ has a $$\subseteqq$$-minimal member. Equivalently, $$T$$ is Tarski-finite iff there is a bijection $$f:T\to \{j\in \Bbb N_0:j for some $$n\in \Bbb N_0$$. A set $$D$$ is Dedekind-infinite iff there exists a bijection $$g:D\to E$$ for some $$E\subsetneqq D$$.' Since the discovery of Forcing, it has been shown that it is consistent with ZF that there exists a set $$B$$ which is neither Tarski-finite nor Dedekind-infinite. So if $$p\in B$$ and $$A=B\setminus\{p\}$$ and $$a=|A|$$ then $$a+1=|B|$$, so $$a\ne a+1$$ because that would imply a bijection $$g:B\to A$$, contrary to $$B$$ not being Dedekind-infinite. So some consequence of AC is needed.