Why is $+1$ necessary in the definition of the $\lt$-minimum element of a set of ordinals

Question

My question pertains to this paragraph in Schimmerling's "Intro to Set Theory," page 34.

If $A$ is a non-empty set of ordinals, then $A$ has an $\lt$-minimum element. To justify the definition, use that fact that $A\subseteq \text{sup}(A)+1$ and $(\text{sup}(A)+1,\lt)$ is a wellordering.

Why do you need the $+1$?

My guess is to deal with, for example, $A=\{1, 2,\dots\}$, where $\text{sup}(A)=\omega$, but $\omega\notin A$.

If this is a correct guess, why is $A$ any more of a subset of $\omega+1$ than $\omega$, and why is $(\omega+1,\lt)$ any more of a wellordering than $(\omega,\lt)$?

Thanks

Answer 1

Suppose $A=\{0,1,2\}$. Then $\sup A=2$, but $A\not\subseteq 2=\{0,1\}$. This is why you need $(\sup A)+1$.

Answer 2

We want an ordinal $B$ such that $A\subseteq B$. So for $A=\{2,17,42\}$, we cannot take $B=43$, we must take $B=42$ so that $42\in B$. Recall that we only know that $x\le \sup A$ for all $x\in A$, we do not know $x<\sup A$.