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Charles C.Pinter - Set theory

Let $a,b,c$ be any cardinal numbers.

Give a counterexample to the rule: $a+b = a+c \implies b=c$

Does there exist a counterexample?

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    Why was this question downvoted?2012-05-11

3 Answers 3

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$\begin{align}&1.\quad\aleph_0=\aleph_0+0=\aleph_0+1\\&2.\quad 0\neq 1\end{align}$

Where $\aleph_0$ is the cardinality of countably infinite sets, e.g. the non-negative integers, $\mathbb N$.

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    @Greg: Indeed. We say that A is a *Dedekind-finite* set if whenever $B\subsetneq A$, |B|<|A|. Equivalently this is to say that |A|<|A|+1. Every finite set is Dedekind-finite, and assuming the axiom of choice the opposite is also true: Dedekind-finite sets are finite sets. However it is consistent that without the axiom of choice there may be infinite Dedekind-finite sets. These sets may be used to construct such counterexamples.2012-05-12
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Here's one: $|\Bbb N| + |\{1\}| =|\Bbb N| + |\{2, 3\}| $, but $|\{1\}| \ne|\{2,3\}|$.

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You will probably learn later that for any infinite cardinal $a$ the equality $a+a=a\cdot a=a$ holds. This implies that for any infinite cardinal $a$ you have a counterexample $a+a=a+0$.

More generally, if $b$, $c$ are infinite cardinals then $b+c=b\cdot c=\max\{b,c\}.$

The proof of these general result is not that simple, it requires axiom of choice. See e.g. the following questions: About a paper of Zermelo and How to prove that from "Every infinite cardinal satisfies $a^2=a$" we can prove that $b+c=bc$ for any two infinite cardinals $b,c$?

However, showing the above result for some special cases, like $a=\aleph_0$ or $a=2^{\aleph_0}$ is not that difficult and it might be a useful exercise for someone learning basics of set theory and cardinal arithmetic.