Ordinal arithmetic $3+\omega^2 = \omega^2$

Question

Which one of the following equalities is false:
a) $2\cdot \omega = 3\cdot\omega$
b) $3+\omega+\omega^2 = \omega+3+\omega^2$
c) $\omega^2 + 3 = 3+\omega^2$
d) $12\cdot(5+\omega)=60 \cdot\omega$

I think the wrong one is c):
In a) and d) both sides are $\omega$. And for b) I found that:
$\omega^2$ can be viewed as $\{0,1,2,\dots,\omega,\omega+1,\dots,\omega\cdot 2,\omega\cdot 2+1,\omega\cdot 2+2,\dots,\omega\cdot 3, \omega\cdot 3+1,\dots\}$.
Thus in b) LHS we have $3+\omega = \omega$ and RHS $3+\omega^2=\omega^2$.
But in c) we get $\omega^2+3\not=\omega^2$.

Is that right?

Answer 1

In c) you have limit ordinal on one side of the equation and non-limit on the other side. The ordinal sum $\alpha+\beta$ is limit if and only if $\beta$ is limit. The same is true for non-limit (successor) ordinals.

So if you have been told that exactly one of them is incorrect, this must be the one.

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The other computations you have given seem to be correct.

a) You have $n\cdot \omega=\omega$ for every $n<\omega$. (This can be visualized as follows: If we replace in the usual ordering of $\omega=\{0,1,2,3,\dots\}$ each number by finitely many copies, we get an order-isomorphic set.)

b) You are correct in saying that $3+\omega=\omega$ and $3+\omega^2=\omega^2$. Then you have $\omega+\omega^2$ on both sides. This can be further simplified as $\omega+\omega^2=1\cdot\omega+\omega\cdot \omega=(1+\omega)\cdot\omega=\omega\cdot\omega=\omega^2$.

d) $5+\omega=\omega$; so you get $12\cdot\omega=60\cdot\omega=\omega$ for the same reasons as in a).