Proof of Proposition 0.17 in Folland wrong?

Question

Proposition 0.17 in Folland's Real Analysis (2e) is

If $X$ and $Y$ are well ordered, then either X is order isomorphic to $Y$, or $X$ is order isomorphic to an initial segment in $Y$, or $Y$ is order isomorphic to an initial segment in $X$.

The proof begins as follows

Consider the set $\mathscr{F}$ of order isomorphisms whose domains are initial segments in $X$ or $X$ itself and whose ranges are initial segments in $Y$ or $Y$ itself. $\mathscr{F}$ is non-empty since the unique $f \colon \{\inf X\} \to \{\inf Y\}$ belongs to $\mathscr{F}$, $\ldots$

Here $\inf X$ denotes the minimal element of $X$. My question is whether it is correct to say that $\{\inf X\}$ is an initial segement.

Folland earlier defines an initial segment as the set

$$I_x = \{y \in X \colon y<x\}$$

for some $x \in X$.

It is not clear to me why there should be an $x \in X$ for which $I_x=\{\inf X\}$

Answer 1

Let $U=\{\,y\in X\mid \inf X<y\,\}$. If $U$ is empty, then $\{\inf X\}=X$; otherwise let $x=\inf U$ and we have $I_x=\{\inf X\}$. So you are right: We'd need to distinguzish the special case that $X$ (or $Y$) is a singleton. To repair that, I'd suggest to notice that $\operatorname{id}\colon\emptyset\to \emptyset$ is in $\mathscr F$ as base case (where $\emptyset =I_{\inf X}$ and $\emptyset=I_{\inf Y}$ are initital segments if $X,Y$ are not empty).

Actually, we already need to make a distiction for the case of $X$ (or $Y$) being empty (and not having an infimum), but that is straightforward.

Answer 2

If $X$ is a non-empty well-ordered set, then every non-maximal element $x$ has an immediate successor, $\inf\{y\in X\mid x<y\}$.

Therefore either $\inf X$ is the maximal element, or it has a successor. In the former case $\{\inf X\}=X$ is an initial segment, otherwise it is the initial segment defined by the successor of $\inf X$.