the smallest transfinite ordinal number

Question

My guess is that $\omega$ is the smallest transfinite ordinal number. To prove this, let $\beta$ be any transfinite ordinal number and let $\operatorname{ord}(B, \le)= \beta$. And I need to show that the usual ordering of natural numbers is order isomorphic to a segment of $B$. But I'm having trouble proving this. How can I show this?

Answer 1

HINT: If an ordinal is infinite, then it has finite initial segments of every possible length.

Answer 2

Here is another answer. I assume from the discussion that "transfinite=infinite". Just remark that if $\alpha$ is finite then $\alpha+1$ is finite. So if $\beta$ is infinite then all finite ordinals are less than $\beta$. This means that $\omega \leq \beta$. $\omega$ being defined as the set of finite ordinals.

Answer 3

Just in case you are interested, it is also proved by Whitehead & Russell at ✳ 263.5 in Principia Mathematica

Which says if $P$ is an infinite well-ordered series, $P$ confined to the terms at a finite distance from the first term of P is a progression($\omega$). It follows that an infinite ordinal is at least as great as $\omega$, and therefore infinite ordinals other than $\omega$ are greater than $\omega$.*

• Whitehead & Russell. Principia Mathematica, Volume 3. Merchant Books, 1913