Is it true that $\lvert\alpha+\beta\rvert=\lvert\alpha\rvert+\lvert\beta\rvert$ for ordinals $\alpha$ and $\beta$?

Question

Suppose $\alpha$ and $\beta$ are ordinals. I was asked to prove that $\lvert\alpha+\beta\rvert=\lvert\alpha\rvert+\lvert\beta\rvert$. However, I think to have found a counterexample for this equality, namely if $\alpha=\omega$ and $\beta=1$ then $$\lvert\alpha+\beta\rvert=\lvert\omega+1\rvert=\lvert\omega\rvert=\omega\neq\omega+1=\lvert\alpha\rvert+\lvert\beta\rvert.$$ Am I doing something wrong or is there a mistake in the exercise?

Answer 1

As I've mentioned in the comments, you are confusing ordinal and cardinal numbers. This is a very common mistake, and it's important to clarify it.

Let me add the subscript $c$ when we talk of cardinals and subscript $o$ when we talk of ordinals. Then we have $\alpha=\omega_o,\beta=1_o$. Then your argument is $$|\alpha+\beta|=|\omega_o+1_o|=|\omega_o|=\omega_c\neq\omega_c+1_c=|\omega_o|+|1_o|=|\alpha|+|\beta|$$

Right now it's easy to see where the mistake is - in the world of cardinal numbers, equality $\omega_c=\omega_c+1_c$ does hold. This is why your counterexample is invalid.

Edit: As Noah points out, the alternative way to look at this is to note that the two ways in which you use $+$ are different - once it is an operation of ordinal sum, the other is cardinal sum.

As Asaf mentions, the safest way to go is to represent cardinals using aleph notation (which has become pretty much a standard in set theory), and leave $\omega$ to only denote ordinals.