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Let be $|M|$ the cardinality of a finite set $M$. This means that there exists some $n\in\mathbb{N}$ and a bijection $f:M\to\{1,\dots,n\}$.

Let's assume that $A\subseteq B$ and both are finite. I was wondering if the two statements

"Cardinality of a finite set is unique"

and

"There exists a bjiection $f:A\to B \iff A=B$"

are equivalent or if at least the first statement implies the second?

Uniqueness of cardinality means: if $f,g$ are two bijections with $f:A\to \{1,\dots,n\}$ and $g:A\to\{1,\dots,m\}$ then $m=n$.

I have gone through some proofs of the second statement, (e.g. https://math.stackexchange.com/a/239567/579544) and it seems that the uniqueness of the cardinality is just a necessary condition but doesn't imply the second statement, does it?

(I hope that this question is not too vague, if so let me know it. )

Philipp
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  • I'm not sure I understand your question. Cardinality, understood as equivalence class under "exists bijection" equivalence, is always unique, regardless of whether a set is finite or not. And your second statement is only true for empty set, unless you assume that $A\subseteq B$ a priori. So, are you asking when finite set is Dedekind finite or vice versa? There is no simple answer to that, it depends on the underlying system of axioms. – freakish Oct 25 '23 at 15:26
  • @freakish, see my edit. I hope it's clear now :) – Philipp Oct 25 '23 at 15:30
  • This kind of uniqueness always holds, the statement is just true. It cannot be otherwise. Regardless of whether set is finite or not. – freakish Oct 25 '23 at 15:31
  • What are A and B? Any finite sets? Two sets can be unequal, and still have a bijection between them, so the existence of a bijection $f: A \rightarrow B$ in almost no context implies $A = B$. – kabel abel Oct 25 '23 at 15:46
  • @freakish, It should be under the assumption of $A\subseteq B$. I put another edit. – Philipp Oct 25 '23 at 15:47
  • @Philipp so what you are asking is whether "being finite" is equivalent to "being Dedekind finite", yes? Here "finite" means "equinumerous with some ${1,\ldots,n}$" while "Dedekind finite" means "if $A\subseteq B$ is equinumerous with $B$ then $A=B$". And as I said in my previous comment: that depends on the exact set theoretic axioms. You can read the wiki I linked in my previous comment. – freakish Oct 25 '23 at 19:22

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