Linear ordering and cardinal numbers - Problem 5.3 Jech's book

Question

I am solving problems of the Jech's book (set theory). I need help to solve problem 5.3: Let $(P, <)$ be a linear ordering and let κ be a cardinal. If every initial segment of $P$ has cardinality $< κ$, then $|P | ≤ κ$. I have the following idea, but it seems it isn't correct path. Let $P$ a linearly ordered set and $\kappa$ a cardinal such that for all $x\in P$, the initial segment $P_{x}=\{y\in P:y<x\}$ is of size less than $\kappa$. If $M$ is a greatest element of $P$, then clearly $|P|=|P_{M}\cup\{M\}|=1+|P_{M}|\leq\kappa$ since $|P_{M}|<\kappa$. Otherwise, let $x_{0}<x_{1}<...x_{\alpha}<...$ an enumeration of P of size $\lambda=|P|$. Then for all $\alpha<\lambda$, $|\alpha|=|P_{{x_{\alpha}}}|<\kappa$ thus $\alpha<\kappa$. Since P does not have a greatest element, $\lambda$ is limit i.e. $\lambda=\sup_{{\alpha<\lambda}}\alpha\leq\kappa$. Again $|P|\leq\kappa$. Do you have a solution?

Answer 1

For $a\in P$ define $P_a:=\{b\in P: b\le a\}.$ So the premise of theorem says that $|P_a|<\kappa$ for all $a\in P.$

Assume, contrary to the conclusion, that $|P|>\kappa,$ and define a sequence $(a_\alpha:\alpha\le\kappa)$ in $P$ by recursion by letting $a_\alpha\in P$ be some element strictly greater than all of $\{a_\beta :\beta <\alpha\}.$ Such an element exists since $\bigcup_{\beta < \alpha}P_{a_\beta}$ is a union of no more than $\kappa$ sets of size less than $\kappa,$ so has size $\le \kappa<|P|.$

But then, since $a_\kappa$ is greater than all of $\{a_\alpha: \alpha < \kappa\}$ we have $|P_{a_\kappa}|\ge \kappa,$ which contradicts the premise.