-1
$\begingroup$

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?

  • 0
    What is an infinite cardinal plus a finite cardinal?2012-05-11
  • 0
    Almost a duplicate: http://math.stackexchange.com/questions/140930/cardinal-number-subtraction2012-05-11
  • 0
    Why was this question downvoted?2012-05-11

3 Answers 3

7

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

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

  • 0
    This is certainly the very simplest counterexample.2012-05-11
  • 0
    or any infinite cardinal2012-05-12
  • 0
    @Greg: Assuming the axiom of choice, yes.2012-05-12
  • 0
    @AsafKaragila: aha, equivalently the well-ordering principle - interesting. Can one construct a counterexample if the axiom of choice isn't assumed?2012-05-12
  • 0
    @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
5

Here's one: $|\Bbb N| + |\{1\}| =|\Bbb N| + |\{2, 3\}| $, but $|\{1\}| \ne|\{2,3\}|$.

1

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.