Counterexample to cancellation law in cardinals addition

Question

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?

Answer 1

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

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

Answer 2

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

Answer 3

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 Simple cardinal arithmetic

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.