# Does an ordinal number $\beta$ such that $\alpha<\beta$ and $\beta<\alpha +1$ exist for some another ordinal number $\alpha$?

My question is simple: is there a pair of ordinals $\alpha,\beta$ such that $\alpha<\beta$ and $\beta<\alpha+1$?

I'm quite sure the answer is no, but I'm a bit afraid of limit ordinals. For me that result follows the fact that given $\alpha$ an ordinal number, $\alpha +1$ is the least element of the set

$$\{\beta:\alpha<\beta \} .$$

However, I haven't found a nice explanation for what a limit ordinal is. I only know what Wiki says: limit ordinal is an ordinal $\omega$ such that there isn't any other ordinal $\alpha$ satisfying $\alpha+1=\omega$.

Hence I'm not totally sure. Because larg ordinals may be very strangers.

What do you think? Thanks.

What do you think?

Recall that $\alpha +1 := \alpha \cup \{\alpha\}$. Now let $\beta < \alpha + 1$, i.e. $\beta \in \alpha + 1$. Then either $\beta \in \alpha$ (and hence $\beta < \alpha$) or $\beta = \alpha$. In particular, there is no ordinal $\beta$ with $\alpha < \beta < \alpha+1$.