Given an infinite set of cardinality $X$, is there a chain in $\mathcal P(X)$ of the same cardinalty of $\mathcal P(X)$?

Solution 1:

Consider the tree $T=2^{\lt\omega_1}$, consisting of all countable ordinal length binary sequences. The number of such sequences is precisely the continuum, since there are $\omega_1$ many levels, each of size continuumn. The lexical order on this tree (or on any two ordinal-length binary sequences), compares them by the order of their least differing bit, placing the shorter one below when one is an initial segment of the other.

Notice that any path $p\in 2^{\omega_1}$ through the tree determines a cut in the lexical order, namely, the collection $X_p$ of $s\in T$ which preceed $p$ in the lexical order. Furthermore, if $p$ preceeds $q$ lexically, then $X_p\subset X_q$, and so we have a chain in $P(T)$ of size $2^{\omega_1}$.

Thus, without any additional assumption, this establishes chains in $P(\mathbb{R})$ of size at least $2^{\omega_1}$.

In particular, if the continuum hypothesis holds, then $|\mathbb{R}|=\omega_1$, and so this shows that there is a chain in $P(\mathbb{R})$ of size $2^{\mathbb{R}}$. So an affirmative answer to your question is a consequence of CH and thus consistent with the axioms of ZFC.

More generally, we obtain a positive answer under the much weaker assumption merely that $2^{\lt\frak{c}}=\frak{c}$, an assumption that follows from but is much weaker than CH. For example, this assumption is compatible with $2^\omega=\aleph_{15}$, if also $2^{\aleph_{14}}=\aleph_{15}$. In this case, the tree $T=2^{\lt\frak{c}}$ has size $\frak{c}$ and $2^{\frak{c}}$ many branches. And so the cuts determined by the lexical predecessors of paths through this tree gives rise to a chain of size $2^{\frak{c}}$ in $P(\frak{c})$, as desired.

Solution 2:

Under the Generalized Continuum Hypothesis, this true by a slight generalization of the Dedekind cut argument:

Assume GCH, and let $\alpha$ be an initial ordinal. (Since GCH implies AC, this is enough to establish the result for an arbitrary infinite set). I will prove that $\mathcal{P} (\alpha)$ contains a chain of cardinality $2^\alpha$.

Let $A=\{X\mid X\subseteq \alpha\}$ with the lexicographic ordering -- that is, $X < Y$ iff $\beta\in Y$ where $\beta$ is the least element on which the two sets differ. (Intuition for $\alpha=\omega$: A is the set of binary fractions between 0 and 1).

Let $B=\bigcup_{\beta<\alpha}\{X \mid X\subseteq \beta\}\subseteq A$. (Intuition: dyadic rationals are countable yet dense).

Because $\alpha \preccurlyeq B = \bigcup_{\beta<\alpha} 2^\beta \preccurlyeq \alpha\times\alpha$, we have $|B|=|\alpha|$. (GCH used for $2^\beta \preccurlyeq \alpha$).

Define an order-embedding $(A,\leq) \to (\mathcal{P}(B),\subseteq)$ by $X \mapsto \{Y\in B \mid Y<X\}$. (Intuition: Dedekind cuts). It is trivial that this mapping is monotonic, and fairly easy to see that it is injective. Thus, the image of $A$ is a chain in $\mathcal{P}(B)\cong\mathcal{P}(\alpha)$ of cardinality the same as $A$, namely $2^\alpha$.

(Adapted from an old Wikipedia refdesk answer of mine.)