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If I have sets [math]A\subseteq B[/math], and I want to show they have same cardinality instead of finding bijection between them could I show simpler bijection to some [math]C[/math] where [math]B\subseteq C[/math], then would [math]A=C \wedge A\subseteq B\subseteq C \implies A=B[/math] or would I need to show something more? Is this clunky?
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>>8710985
That is enough.
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sounds fine, certainly |A| <= |B| along with |B|<= |C| and |A|=|C| gives |A|=|B| since |A| <= |B| <= |C| =|A| implies it right away
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>>8710985
All you need is an injection B->A, your stuff would show it exists.
Example: 2Z and 4Z, 4Z is a subset of 2Z, and 4Z has the same cardinality as Z. Since 2Z is a subset of Z, you have card(4Z)<=card(2Z)<=card(Z)=card(4Z)
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>>8710985
You're using the Schroder-Bernstein theorem here, which constructs a bijection between sets A and B with injections going both ways. It is also true that sets with surjections going both ways have the same cardinality, although this requires the axiom of choice.
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>>8711625
>this requires the axiom of choice
literally shaking rn
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