$(\kappa \times 2,<_{lex}) \cong (\kappa,\in )$ for cardinal numbes $\kappa$.

A cardinal number $$\alpha$$ is a ordinal number such that $$\forall \beta < \alpha : \beta \not\approx \alpha$$. Now I want to show that for a cardinal number $$\kappa$$ the set $$\kappa \times 2$$ with the lexicographic order $$<_{lex}$$ defined by $$(a,b) <_{lex} (a',b') :\iff a < a' \lor (a' = a \land b < b')$$ is order isomorph to $$\kappa$$ with the well order $$\in$$.

So far I was able to show that $$\forall \lambda < \kappa : \kappa \times 2 \approx \lambda$$ and for infinite ordinal numbers $$\alpha$$ one has $$\kappa \times 2 \approx \alpha$$.

Edit: By transfinite induction on $$\kappa$$. We know $$\omega \times 2 \approx \omega$$ and $$|\omega \times 2| = \omega$$. Thus we can assume that it holds for smaller infinite cardinals $$\lambda < \kappa$$ und prove it for $$\kappa$$. My Idea is to note that $$<_{lex}$$ is a well ordering on $$\kappa \times 2$$ and starts out by $$$$(0,0) <_{lex} (1,0) <_{lex} (2,0) <_{lex} (3,0) <_{lex} \ldots <_{lex} (\omega,0) <_{lex} (\omega,1).$$$$ Now each $$(a,b) \in \kappa \times 2$$ has $$|\max(a,b) + 1 \times \max(a,b) + 1| < \kappa$$ many $$<_{lex}$$ predecessors and $$otp((\kappa \times 2)) = \kappa$$ hence there must be a order isomorphism between $$(\kappa \times 2, <_{lex})$$ and $$(\kappa, \in)$$.

• What makes you think this is true at all? Commented Jan 24, 2023 at 19:50
• At least it is true for $\omega_0$ since one can construct a chain $\lbrace (1,0),(1,1),(2,0), (2,1), \ldots \rbrace$ in increasing order which is order isomorphic to $\omega_0$. Therefore $2 \times \omega_0$ is order isomorphic to $\omega_0$ and more general $n \times \omega_0$ is order isomorphic to $\omega_0$. So my thought to show this was to use a similar approach but I am not sure how.
– Orb
Commented Jan 24, 2023 at 20:01
• I think you mean to be considering $\kappa\times 2$ rather than $2\times\kappa$. Commented Jan 24, 2023 at 20:46
• Sorry for nitpicking: $\kappa$ has to be infinite or $0$.
– Ulli
Commented Jan 25, 2023 at 13:18

What you want to show is that if $$\kappa$$ is a cardinal then it is the unique order type of a well-order that has cardinality $$\kappa$$ and every proper initial segment has size $$<\kappa$$.
So, for example, $$\omega$$ is the unique well-ordering that is countable and has every proper initial segment finite. And $$\omega_1$$ is the unique order type of a well-order which is uncountable and every proper initial segment is countable.
So, to prove that $$\kappa\times2$$ is isomorphic to $$\kappa$$, it is enough to show that any proper initial segment has size $$<\kappa$$, as very easily $$\kappa\times2$$ has cardinality $$\kappa$$.
For this, show that if $$A\subseteq\kappa\times2$$ is a proper initial segment, then there is some $$\gamma<\kappa$$ such that $$A\subseteq\gamma\times2$$.