Do limit ordinal always have a preceding limit ordinal

Question

I am trying to prove that a non zero limit ordinal $\gamma$ always has a preceding limit ordinal (counting zero as a limit ordinal).

I am not sure if this is true but if it is $\bigcup\{\delta\mid\delta < \gamma \text{ and } \delta \text{ is a limit ordinal} \} $ would be the candidate.

I can prove a union of limit ordinals is a limit ordinal. It's proving the above limit ordinal is less than $\gamma$ that is unclear to me.

Answer 1

The statement you are trying to prove is not true. If $\lambda$ is a (nonzero) limit of limit ordinals, then - by definition - there is no greatest limit ordinal $<\lambda$.

The first limit-of-limits is $\omega^2$: we have $\omega^2=\sup\{\omega,\omega\cdot 2,\omega\cdot 3,...\}$ and clearly each $\omega\cdot n$ is a limit ordinal. Similarly, there are ordinals which are limits-of-limits-of-limits such as $\omega^3=\sup\{\omega^2,\omega^2\cdot 2,\omega^2\cdot 3,...\}$ - or $\omega^\omega$, or $\epsilon_0$, or $\omega_1$, or etc.

Answer 2

The very first limit ordinal is $\omega$, it's the ordinal that's the union of all the natural numbers. It clearly has no immediate predecessor. As per the comment, that's what makes these limit ordinals.