# Which ordinals can be order-embedded in $2^\kappa$ for a given infinite cardinal $\kappa$?

Let $$\kappa$$ be an infinite cardinal. The set $$2^\kappa=\{0,1\}^\kappa$$ is given the lexicographically order in the usual way ($$f if $$f(\alpha) at the first position where the functions differ). I am wondering which ordinals can be embedded (as a linear order) into $$2^\kappa$$.

The cardinal $$\kappa$$ viewed as an ordered set embeds into $$2^\kappa$$ by sending each ordinal $$\gamma<\kappa$$ to its characteristic function in $$\kappa$$. So any ordinal $$\alpha\le\kappa$$ embeds into $$2^\kappa$$. On the other hand, Jech's Set Theory Lemma 9.5 says:

The lexicographically ordered set $$\{0,1\}^\kappa$$ has no increasing or decreasing $$\kappa^+$$-sequence.

So $$\kappa^+$$ and any larger ordinals do not embed into $$2^\kappa$$.

What can we say about the ordinals $$\alpha$$ with $$\kappa<\alpha<\kappa^+$$?

For example, for $$\kappa=\aleph_0$$, any countable ordinal embeds into $$2^{\aleph_0}$$ (because every such ordinal embeds into $$(\mathbb{Q},<)$$, which embeds into $$(\mathbb{R},<)$$, which embeds into $$2^\omega$$).

For $$\kappa=\aleph_1$$: Does $$\omega_1+1=\{\alpha: 0\le\alpha\le\omega_1\}$$ embed into $$2^{\omega_1}$$? Or higher ordinals less than $$\omega_2$$?

Anything known in general? The answer possibly depends on some set theoretic assumptions (CH, etc). Any references would be appreciated.

Added: I think we can always embed $$\kappa+1$$, which is the same as $$\kappa$$ plus a extra maximum point. The reason is that $$2^\kappa$$ also has a maximum point, namely the function identically equal to $$1$$, and $$\kappa$$ has no maximum, so just extend the embedding of $$\kappa$$ appropriately. More generally, I see how to embed any ordinal less than $$\kappa\cdot\kappa$$ (ordinal product), but not sure in general.

Any ordinal $$\alpha<\kappa^+$$ can be embedded in $$2^\kappa$$ with respect to the inclusion order (and thus also with respect to the lexicographic order which is stronger). To prove this, just note that $$\alpha$$ is isomorphic to its set of initial segments, ordered by inclusion. That is, $$\alpha$$ embeds in $$2^\alpha$$ with the inclusion order. Since $$\alpha<\kappa^+$$, there exists an injection $$\alpha\to\kappa$$, which gives an embedding of $$2^\alpha$$ into $$2^\kappa$$ with the inclusion order.
(Similarly, every partial order of cardinality $$\leq\kappa$$ embeds in $$2^\kappa$$ with respect to the inclusion order, by sending each point to the set of elements less than or equal to it. For a total order, this implies it will then also embed in any total order on $$2^\kappa$$ which contains the inclusion order.)