TITLE: Cardinality of a some linear ordering is at most that of a given cardinal?
QUESTION [8 upvotes]: This is an intuitive idea that I've used for a while, but don't know how to explain formally.
Suppose $(A,\prec)$ is some linear ordering, and each initial segment of $A$ has cardinality strictly smaller than $\kappa$ for some cardinal $\kappa$. Then $|A|\leq\kappa$.
This makes sense, since you can just take bigger and bigger initial segments whose cardinalities "approach" $\kappa$. But more formally, why is $|A|\leq\kappa$?

REPLY [2 votes]: The statement is true.
Suppose that $|A|>\kappa$. For each $x\in A$, $A$ is the disjoint union of $(\leftarrow,x)$, $\{x\}$, and $(x,\to)$, and $|\{x\}|,|(\leftarrow,x)|<\kappa$, so $|(x,\to)|>\kappa$. In other words, every ‘tail’ of $A$ has cardinality greater than $\kappa$. I’ll use this to construct a strictly increasing function $\varphi:\kappa^+\to A$.
Let $\varphi(0)\in A$ be arbitrary. Suppose that $\eta<\kappa^+$, and I’ve already chosen $\varphi(\xi)\in A$ for each $\xi<\eta$ in such a way that $\varphi\upharpoonright\eta$ is strictly increasing. Let $$P_\eta=\bigcup_{\xi<\eta}\big(\leftarrow,\varphi(\xi)\big)\;;$$ by hypothesis $\left|\big(\leftarrow,\varphi(\xi)\big)\right|<\kappa$ for each $\xi<\eta$, and $|\eta|\le\kappa$, so $|P_\eta|\le\kappa$. Thus, $A\setminus P_\eta\ne\varnothing$, and we can choose $\varphi(\eta)\in A\setminus P_\eta$. Clearly $\varphi(\eta)>\varphi(\xi)$ for $\xi<\eta$, so the induction goes through to $\kappa^+$.
But this is impossible, because $\varphi(\kappa)$ has at least $\kappa$ predecessors in $\langle A,\prec\rangle$, namely, $\{\varphi(\xi):\xi<\kappa\}$.