Prove or disprove:
If $A\times B\sim A\times C$, then $B\sim C$.
("$\sim$": "numerically equivalent" / "has the same cardinality as")
What bijection/counterexample should I use to prove/disprove it?
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Hint: Do you know of a non-empty set $A$ with $|A\times A| = |A|$? – Tobias Kildetoft Jan 08 '15 at 12:38
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@Tobias: Why non-empty? (This question is usually given before a proof that $|\Bbb N|=|\Bbb{N\times N}|$.) – Asaf Karagila Jan 08 '15 at 12:38
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2Proof requires further assuming that $A,B,C$ are finite. – hardmath Jan 08 '15 at 12:38
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@hardmath: Additional assumptions are required. – Asaf Karagila Jan 08 '15 at 12:39
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It was only stated that A, B and C are sets. Though you may be right. @hardmath – Jan 08 '15 at 12:40
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@AsafKaragila: Okay, finiteness and $A$ nonempty are sufficient. – hardmath Jan 08 '15 at 12:40
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@AsafKaragila Hmm, not sure how I managed to mention non-empty and not think one step further. – Tobias Kildetoft Jan 08 '15 at 12:40
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Assuming that $|A|\leq |B|$ and $|A|\leq |C|$ should also be sufficient. – Arthur Jan 08 '15 at 12:41
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@AsafKaragila: If we assume the sets are non-empty and $X \sim Y$ means there is a bijection $\phi : X \to Y,$ won't it be sufficient? – Krish Jan 08 '15 at 13:05
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@Krish: You will also need some finiteness condition. – Asaf Karagila Jan 08 '15 at 13:22
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@Guest What is the exact question: the one from the title ($;|A \times B| = |A \times C| \implies |B|=|C|;$), or the one from the question itself ($;A \times B \sim A \times C \implies B \sim C;$)? The former is only meaningful for finite sets, or if you're clear about what kind of arithmetic you use for infinities. The latter works also for infinite sets, and is usually bijection. – MarnixKlooster ReinstateMonica Jan 09 '15 at 17:51
2 Answers
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HINT: You are asked to divide $|A|\cdot|B|=|A|\cdot|C|$ by $|A|$. Can you think of $A$ such that $|A|$ is a number that cannot be reduced from both sides of the equation?
Asaf Karagila
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Consider $A=\mathbb{N}$. Now take $B$ and $C$ any different, finite, non-empty sets. Actually it seems that you can take $A$ any infinite set, but for this one I'm not sure.
Mihail
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