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For me every connected set has at least two elements.

Connected subsets of the real line have non-empty interiors. I am curious if the same is true for connected subsets of any connected linearly ordered topological space (the topology comes from the order).

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Yes. In fact, the space need not be connected, as all connected subsets of a linearly ordered topological space are intervals, and in a linearly ordered space connected intervals have nonempty interior. Let $X$ be such a space. To prove the first statement, assume $S$ is connected but not an interval, so we have some $a,b\in S$ and $x\in X$ such that $a

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    The argument isn’t quite complete: $(a,b)$ could be empty. But in that case $\{a\}$ and $\{b\}$ separate $[a,b]$.2011-12-26
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    @Brian Thank you for the correction, I will edit to reflect that.2011-12-26
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    Thank you. By the way, assuming the Souslin line exists, must its density character be $\omega_1$?2011-12-26
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    @S.Serva I'm not familiar with density character, so I'm afraid I can't help you there.2011-12-26
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    @S.Serva: Yes. Every Suslin line contains an Aronszajn line as a dense subset, and every Aronszajn line has cardinality $\omega_1$.2011-12-26
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    Alex, your edit introduced a new error: the closure of $I$ need not be of the form $[a,b]$. I’ve taken the liberty of fixing the argument; please check to be sure that the changes are acceptable to you.2011-12-26
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    @Brian: Yes, thanks for the second correction.2011-12-26
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    @S.Serva in fact, for all ordered spaces $d(X) \le c(X)^{+}$, so density of $X$ and cellularity of $X$ are at most a successor step away from each other.2011-12-27