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I apologize for this naive question.

Let $\eta$ be an ordinal. Isn't the supremum of $\eta$ just $\eta+1$? If this is true, the supremum is only necessary if you conisder sets of ordinals.

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2 Answers 2

When you consider $\eta$ a set of ordinals, its supremum is $\eta$ itself if $\eta$ is a limit ordinal, and otherwise it is the predecessor of $\eta$.

The supremum of the singleton set $\{\eta\}$ is of course $\eta$ itself.

However, when dealing with sets of ordinals, it is often more useful to consider the least ordinal strictly greater than every element of the set. This operation maps every ordinal (considered as a subset of ON) to itself.

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There are two notions of supremum in sets ordinals, let $A$ be a set of ordinals:

  • $\sup^+(A)=\sup\{\alpha+1\mid\alpha\in A\}$,
  • $\sup(A) =\sup\{\alpha\mid\alpha\in A\}$.

Since most of the time we care about supremum below limit ordinals (eg. $A=\omega$) the notions coincide. If $A=\{\alpha\}$ then indeed $\sup(A)=\alpha$ and $\sup^+(A)=\alpha+1$.

The reason there are two notions is that $\sup^+(A)$ is defined as $\min\{\alpha\in\mathrm{Ord}\mid A\subseteq\alpha\}$ and $\sup(A)=\bigcup A$.

Both of these notions are useful, and it is easy to see that if $A$ has no maximal element then these indeed coincide. However the distinction can be useful in successor ordinals from time to time.

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