If $\gamma$ is an uncountable ordinal, then $\gamma$ contains uncountably many successor ordinals.

Question

Suppose $\gamma$ is an uncountable ordinal, i.e. $\gamma$ has uncountably many elements. We write $$ \gamma = \{0\} \cup\{\alpha \in \gamma: \alpha \text{ is a successor ordinal}\} \cup\{\alpha \in \gamma: \alpha \text{ is a limit ordinal}\} =: Z \cup S \cup L. $$

Since $\gamma$ is uncountable, we must have at least one of $Z,S,L$ uncountable. I would like to show that no matter what, $S$ is uncountable.

Clearly $Z$ cannot be uncountable, and if $S$ is uncountable we are done, so we consider the case where $L$ is uncountable. That is, there are uncountably many limit ordinals in $\gamma$. From this can we show that $S$ must also be uncountable?

Answer 1

Consider the ordinals $\alpha+1$ where $\alpha$ is a limit ordinal in $\gamma$. How many of those are in $\gamma$?

Answer 2

Suppose that $\gamma$ is a limit ordinal, if not then it is $\gamma'+n$ for some $n\in\omega$, and $\gamma'$ is uncountable, the following argument works for $\gamma'$ and proves the claim for $\gamma$. The map $\alpha\mapsto\alpha+1$ is an injection from $\gamma$ into $S$. Therefore $S$ must be uncountable.

Also note that $\delta+n\mapsto\delta$ where $\delta\in L$ and $n\in\omega$ is a surjection from $S\setminus\omega$ onto $L$, and each fiber is countable. Therefore $L$ is also uncountable.