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Let $\lambda$ be an uncountable cardinal and let $X=\{0,1\}^\lambda$ be endowed with the product topology ( $\{0,1\}$ is discrete). Is there an uncountable chain (with respect to inclusion) of clopen subsets of $X$?

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It took a while, but I finally got the general result (barring mental hiccups).

Suppose that $\mathscr{C}$ is a chain of clopen sets in $X=\{0,1\}^\lambda$ of cardinality $\omega_1$. The compactness of $X$ implies that the clopen sets are precisely the unions of finitely many basic clopen sets of the form $B(\varphi)=\{x\in X:x\upharpoonright\operatorname{dom}\varphi=\varphi\}$, where $\varphi$ is a finite partial function from $\lambda$ to $\{0,1\}$. By passing to a subchain if necessary, we may assume there is an $n\in\omega$ such that each $C\in\mathscr{C}\,$ is the union of basic clopen sets $B(\varphi^C_1), B(\varphi^C_2), \dots,B(\varphi^C_n)$. For each $C\in\mathscr{C}\,$ let $$D(C)=\bigcup_{k=1}^n\operatorname{dom}\,\varphi^C_k\;;$$ by the delta-system lemma we may assume that $\{D(C):C\in\mathscr{C}\,\}$ is a delta-system with common part $D$. We may further assume that for $k\in[n]$ there are $D_k\subseteq D$ and $\psi_k:D_k\to\{0,1\}$ such that $$D=D_1\cup\dots\cup D_n$$ and for all $C\in\mathscr{C}$ and $k\in[n]$, $$D\cap\operatorname{dom}\varphi^C_k=D_k\text{ and }\varphi^C_k\upharpoonright D_k=\psi_k\;.$$

For $k\in[n]$ let $D^C_k=\operatorname{dom}\varphi^C_k\setminus D_k$ and $\sigma^C_k=\varphi^C_k\upharpoonright D^C_k$. Since there are only countably many possibilities, we may assume that for all $A,C\in\mathscr{C}$ the set systems $\left\langle D^A_1,\dots,D^A_n\right\rangle$ and $\left\langle D^C_1,\dots,D^C_n\right\rangle$ are isomorphic in the sense that there is a bijection $$\pi_{A,C}:D(A)\setminus D\to D(C)\setminus D$$ such that for any $S\subseteq [n]$, $$\pi_{A,C}\left[\bigcap\{D^A_k:k\in S\}\right]=\bigcap\{D^C_k:k\in S\}\;,$$ and we may further assume that $\sigma^A_k=\sigma^C_k\circ\pi_{A,C}$ for $k\in[n]$. Then we can extend $\pi_{A,C}$ to a permutation $\hat\pi_{A,C}$ of $\lambda$:

$$\hat\pi_{A,C}(\xi)=\begin{cases} \pi_{A,C}(\xi),&\text{if }\xi\in D(A)\setminus D\\\\ \pi_{A,C}^{-1}(\xi),&\text{if }\xi\in D(C)\setminus D\\\\ \xi,&\text{otherwise}\;. \end{cases}$$

Finally, let $h_{A,C}:X\to X:x\mapsto x\circ\hat\pi_{A,C}$; then $h_{A,C}$ is an autohomeomorphism of $X$ such that $h_{A,C}[A]=C$ and $h_{A,C}[C]=A$. But then $x\in C\setminus A$ iff $h_{A,C}(x)\in A\setminus C$, so $A\subseteq C$ iff $A=C$, and $\mathscr{C}$ cannot be an uncountable chain after all.

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Generally, "chain" just means totally ordered, but perhaps not well ordered. –  GEdgar Dec 14 '11 at 4:41
    
@GEdgar: I’m aware of that; it’s one reason I was so careful to specify my interpretation. (I’m still thinking about the general case.) –  Brian M. Scott Dec 14 '11 at 5:02
    
Thank you, but I don't ask about well-ordered chains. Btw. Your claim follows from the fact that $\{0,1\}^\lambda$ is ccc. –  Bakcyl Dec 14 '11 at 11:10
    
@Bakcyl: It wasn’t clear from your question whether or not you knew that chain doesn’t imply well-ordered. A lot of questions come from students who aren’t careful about terminology. As for the rest, I know: in effect I was proving that $\{0,1\}^\lambda$ is ccc. Again, I couldn’t tell from the question whether you were even familiar with the concept of countable cellularity, so I thought it better simply to give the argument. –  Brian M. Scott Dec 14 '11 at 19:10
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