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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 but $x\notin S$. Then the sets $(-\infty, x)\cap S$ and $(x,\infty)\cap S$ are open in $S$, nonempty, and their union covers $S$, contradicting the fact that $S$ is connected. To prove the second, let $I$ be a connected interval in $X$ with at least two points $a,b\in I$, chosen so that $a. Then the open set $(a,b)$ is contained in $I$, and the set $(a,b)$ cannot be empty, as otherwise we would have that $I = \big((-\infty,b)\cap I\big)\cup\big((a,\infty)\cap I\big)$ is disconnected as these are disjoint nonempty relatively open sets. Thus $I$ has nonempty interior.

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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