FALSE: every uncountable subset of $\mathbb{R}$ contains an interval

FALSE. Every uncountable subset $X \subseteq \mathbb{R}$ (Finite, countably infinite, countable, uncountable) contains a nondegenerate interval: there are $a < b$ in $\mathbb{R}$ with $(a,b) \subseteq X$.

The claim is plausible because an uncountable set is, in a rough sense, large, and the intervals are the obvious large subsets of $\mathbb{R}$. But size in the sense of cardinality says nothing about how a set sits inside $\mathbb{R}$: a set can be uncountable and still meet every interval in a set with holes. The irrationals are the standard witness, and the Cantor set, once measure and topology are available, is a starker one.

• The counterexample is as strong as possible in one direction: $X_0$ misses no interval either, so it is dense and yet contains no interval. That $X_0$ meets every $(a,b)$ with $a < b$ needs no new input, only what is already on this page: were $(a,b) \cap X_0$ empty we would have $(a,b) \subseteq \mathbb{Q}_{\mathbb{R}}$, and $\mathbb{Q}_{\mathbb{R}} = \iota[\mathbb{Q}]$ is at most countable, being a bijective image of $\mathbb{Q}$ ($\mathbb{Q}$ is countably infinite, The unique embedding of ℚ into an ordered field), so $(a,b)$ would be at most countable (Every subset of an at most countable set is at most countable), which it is not, by the next remark. Density and containing an interval are unrelated properties.

• Every nondegenerate interval is uncountable, open as well as closed (Every nondegenerate interval of $\mathbb{R}$ is uncountable). The open form is the one the remarks on either side of this one need, and the corollary states it outright, so nothing has to be transported here from the closed case to the open one. It is proved by re-running the nested-interval construction of $\mathbb{R}$ is uncountable (Cantor's nested intervals, 1874) seeded at the middle third of $(a,b)$, which is what places the point that construction produces strictly inside $(a,b)$ rather than merely in $[a,b]$; the density of $\mathbb{Q}_{\mathbb{R}}$ recorded in [L2] is not needed for it.

• The converse implication is true and trivial: a nondegenerate interval is uncountable, by the previous remark, so "contains an interval" implies "uncountable" (Every subset of an at most countable set is at most countable again, applied to the interval inside the set). Only the direction claimed above fails.

• A cardinality assumption cannot be repaired into a topological conclusion. The Cantor set is uncountable, closed, and contains no interval; it also has measure zero, so it is small in a second, independent sense. Neither notion is developed here, and neither is needed: the irrationals already settle the question.