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My book states this as obvious, but it isn't so trivial to me. thanks

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Can you come up with a bijection between $\alpha$ and $\alpha + 1$, where $\alpha$ is an infinite ordinal? – Alexander Thumm Feb 16 '13 at 16:22

Recall that a cardinal is an initial ordinal, an ordinal which cannot be put in bijection into a smaller ordinal. Thanks to the Cantor-Bernstein theorem it is enough to show that there is no injection into a smaller ordinal instead.

If an infinite ordinal $\alpha$ is not a limit ordinal then there is some $\beta$ such that $\alpha=\beta\cup\{\beta\}$.

The map $g\colon\alpha\to\beta$ defined as: $$g(x)=\begin{cases} 0 & x=\beta\\x+1 & x<\omega\\x & \text{otherwise}\end{cases}$$ is injective from $\alpha$ into a smaller ordinal, and therefore $\alpha$ cannot be an initial ordinal, that is a cardinal.

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They are ordinals by definition, because we choose to use initial ordinals as the canonical representative of each class of sets-with-the-same-cardinality. This choice could in principle have been made differently, but it turns out to be a useful convention (though it only works in when we have the Axiom of Choice).

They are limit ordinals, because (except for 0) limit ordinals are precisely the ordinals that are not successor ordinals. And an infinite successor ordinal always has the same cardinality as the one it is a successor of (you can make a bijection that moves the last element away into the initial $\omega$ segment with the standard Hilberts-Hotel trick), so a successor ordinal is never the first ordinal with a given cardinality.

(On the other hand, beware that not all limit ordinals are cardinals).

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