Is this a correct statement: $\sup \omega = \omega + 1$? I'm sorry if this is very obvious, but I am re-reading Rudin's Real Analysis, and thinking of $\omega$. Would it be correct to say that the supremum of $\omega$ is $\omega + 1$?  And if so, how can I say $\omega$ is an unbounded set while also saying that it has a least upper bound in the ordinals? What's the best way to say this?
Edit:  
$\omega = \{0, 1, 2, \dots \}$
$\omega + 1 = \{0, 1, 2, \dots, \omega \}$
 A: The supremum of a subset $\ S\ $ of an ordered set $\ T\ $ is, by definition, an element $\ t\in T\ $ such that $\ s\le t\ $ for all $\ s\in S\ $ and there does not exist $\ u\in T\ $ such that $\ s\le u < t\ $ for all $\ s\in S\ $.
If $\ S=T=\omega\ $, then there is no $\ t\ $ in $\ t\in T\ $ such that $\ s\le t\ $ for all $\ s\in S\ $, so we can indeed say that $\ \omega = S\ $ is "unbounded" in this case. 
If $\ S=\omega\ $ and $\ T\ $ is a set of ordinals with $\ \omega\cup\{\omega,\omega+1\}\subseteq$$ T,$ however, then $\ s\le\omega\ $ for all $\ s\in S\ $, and there is no $\ t\in T\ $ such that $\ s\le t<\omega\ $ for all $\ s\in  S\ $, so $\ \sup S=\sup\omega = \omega\ $ (not $\ \omega+1\ $) in this case.
In fact, if $\ T\ $ is any ordinal, and $\ \alpha\in T\ $, then $\ \sup\alpha=\alpha\ $ if and only if $\ \alpha\ $ is a limit ordinal. If $\ \alpha=\beta+1\ $ is a successor ordinal, then $\ \sup\alpha=\beta\ $.
A: No, $\sup \omega=\omega$; firstly $\omega$ is an upperbound for $\omega$: 
$$\forall n \in \omega: n \le \omega $$
which is clear as $\alpha < \beta$ iff $\alpha \in \beta$ (iff $\alpha \subsetneq \beta$) for ordinals generally, so $\alpha \le \beta$ iff $\alpha=\beta$ or $\alpha \in \beta$ (or $\alpha \subseteq \beta$ too).
and if the ordinal $\beta$ is an upper bound for $\omega$, so
$$\forall n \in \omega: n \le \beta $$ and note that $\beta \notin \omega$, or otherwise $\beta+1 \in \omega$ and we would have $\beta+1 \le \beta$, contradiction. So $n=\beta$ never holds in the $\le$ clause and
$$\forall n \in \omega: n \in \beta \text{ or: } \omega \subseteq \beta \text{ so } \omega \le \beta$$ making $\omega$ the smallest upperbound, hence the sup for $\omega$.
