$m$ is a cardinal. How to prove if $\omega \le m^2$ then $\omega \le m$ without axiom of choice?
I want to construct an injection, but I am not sure how to.
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Xiang
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If you have the axiom of choice, https://math.stackexchange.com/questions/2901201/a-cardinality-number-to-the-square-is-itself – Eric Towers Dec 05 '21 at 04:46
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Thank you for comment! I want to prove it without AC. – Xiang Dec 05 '21 at 04:53
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The word "than" is the wrong word to use. It should instead be "then". – Geoffrey Trang Dec 05 '21 at 04:56
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is it true? what if we consider $\omega = m^2$ then the statement will prove AC – YOu will not know Dec 05 '21 at 06:05
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@AdityaDwivedi That's incorrect - you don't need AC to prove $\omega=\omega^2$ (as cardinals). Indeed, the OP's statement is provable without choice as Geoffrey Trang's answer below shows. – Noah Schweber Dec 05 '21 at 06:41
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@Aditya In case it was missed, note $\omega$ isn't an arbitrary cardinal, which is why we cannot deduce AC from this. – Brian Moehring Dec 05 '21 at 07:05
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Given an infinite sequence $((a_i,b_i))_{i \ge 0}$ of distinct elements of $X \times X$ (where $X$ is a set of cardinality $m$), form the sequence $(a_0,b_0,a_1,b_1,a_2,b_2,...)$. This sequence is an infinite sequence, which may, however, contain some repetitions. Let $Y$ be the set formed by the terms of this sequence (without repetitions). Then, $Y$ must be infinite, because otherwise, the sequence $((a_i,b_i))_{i \ge 0}$ would contain infinitely many distinct elements of the finite set $Y \times Y$, which is a contradiction. So, removing duplicates still gives an infinite sequence, of distinct elements of $X$ this time.
Geoffrey Trang
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