I am trying to understand Theorem 1.7.4 of Devlin's Joy of Sets. The Theorem states:
There is no isomorphism of $X$ onto a segment of $X$ (supposing $X$ is a woset).
The difficulty I am having is that, on the previous page, it is established that a subset, say $Y$, of a set $X$ with an isomorphism $f: X \cong Y$ implies that $\forall x \in X, x \le f(x)$ (Theorem 1.7.2).
Is a segment not also a subset? I understand that Theorem 1.7.2 is conditional on an order isomorphism existing, as it is a premise of the proof. I understand the argument for Theorem 1.7.4 but surely a segment is a subset of a set? It seems like you can construct an isomorphism between $X$ and a segment.
Suppose, $X =$ {$1, 2, 3, 4, 5$} and $X_4 =$ {$1, 2, 3$} is a segment of $X$. Both $X$ and $X_4$ are wosets under the ($\le$) relation.
Now, define $f: X \to X_a$ as a function that maps $1 \to 1, 2 \to 2,$ and $3 \to 3$. Here, $f$ is a bijective function, and it meets the definition of an order isomorphism: $x \lt y \to f(x) \lt ' f(y)$ (in my example the relation between the two sets if the same, so $\lt = \lt '$).
My question then, is why a set cannot have an isomorphism to a segment. Any help would be greatly appreciated!