Ordinals multiplication

44 Views Asked by At

Given that $\gamma \leq m$ prove that $n\cdot S(\gamma) \subset n\cdot S(m)$ where $n,m ,\gamma$ are finite ordinal, meaning elements of $\omega$ and that $S$ is the successor function ?

I used induction but it does not help !

Any simpler proof, thanks in advance.

1

There are 1 best solutions below

0
On BEST ANSWER

If a,b are finite ordinals and a <= b, then na <= nb.
Prove that by induction on n.
By transitivity na $\subseteq$ nb.

Apply that to your problem by proving
$\gamma$ <= m implies S($\gamma$) <= S(m).