Let $\kappa$ be an infinite cardinal. The set $2^\kappa=\{0,1\}^\kappa$ is given the lexicographically order in the usual way ($f<g$ if $f(\alpha)<g(\alpha)$ 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.


1 Answer 1


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.)

  • $\begingroup$ Very nice answer. Thank you. $\endgroup$
    – PatrickR
    Dec 25, 2022 at 0:03

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