Example of an uncountable subset of $[0,1]$ with no uncountable closed subsets Question: Find an uncountable set $A \subseteq [0,1]$ s.t. for every uncountable subset $B$ of $A$, $B$ is not closed.
Solution: My guess would be that $[0,1] \setminus \mathbb{Q}$ would be such a set, but I cant seem to prove it. Any help would be great.
 A: Recall that a set of reals is called "perfect" if it is closed, nonempty, and without isolated points.  Three standard facts about perfect sets (not requiring the Axiom of Choice) are: (a) every uncountable closed set of reals is the union of a perfect set and a countable set ("Cantor-Bendixson"), (b) every perfect set of reals has the cardinality of the continuum, and (c) there are continuum-many perfect sets.
Using the Axiom of Choice, let $(P_\alpha)_{\alpha<\mathfrak{c}}$ be an enumeration of perfect sets (where $\mathfrak{c}$ is the cardinal of the continuum, as per (c) above).  Define two sequences $(x_\alpha)_{\alpha<\mathfrak{c}}$ and $(y_\alpha)_{\alpha<\mathfrak{c}}$ of real numbers by induction (and again using Choice): assuming the $x_\beta$ and $y_\beta$ for $\beta<\alpha$ have already been defined, let $x_\alpha$ and $y_\alpha$ be elements of $P_\alpha$ distinct from each other and distinct from all all $x_\beta$ and $y_\beta$ for $\beta<\alpha$ (this is possible because $P_\alpha$ has cardinality greater than the set of $x_\beta$ and $y_\beta$, per (b) above).  Let $X$ be the set of the $x_\alpha$.  So $X$ has the cardinality of the continuum.  But if $X$ contained an uncountable closed set, then per (a) above it would contain a perfect set $P$, say $P = P_\alpha$, so it would contain $y_\alpha$, a contradiction since the $x_\alpha$ and $y_\alpha$ are disjoint.
Assuming the consistency of an inaccessible cardinal, the Axiom of Choice cannot be dispensed with: it is consistent (relative to the existence of an inaccessible cardinal) that the Axiom of Dependent Choices holds and that every uncountable set of reals has a perfect subset.  The latter statement is, also, a consequence of the Axiom of Determinacy, which is also known to be consistent with Dependent Choices, subject to the consistency of hugely larger cardinals.
