The part of the question about the continuum hypothesis (CH)continuum hypothesis (CH) seems confused: without assuming (CH) (but assuming axiom of choice so that cardinals work as they should), aleph_1$\aleph_1$ is by definition the least uncountable cardinal. (The continuum hypothesis asserts that aleph_1 = 2^{aleph_0}$\aleph_1 = 2^{\aleph_0}$.)
Let X$X$ be a metric space, and let x in X$x \in X$. Then it follows immediately from the definition -- see e.g.
http://en.wikipedia.org/wiki/Hausdorff_measure
that for any d > 0$d > 0$, the d$d$-dimensional Hausdorff measure H_d({x})$H_d(\{x\})$ is equal to zero. (This is just because a point can be covered by a single ball with arbitrarily small diameter.) Since H_d$H_d$ is a measure, it is countably additive: thus H_d(S) = 0$H_d(S) = 0$ for any countable set S$S$. If H_d(S) = 0$H_d(S) = 0$, then the Hausdorff dimension of S$S$ is at most d$d$, so H_d(S) = 0$H_d(S) = 0$ for all positive d$d$ implies that the Hausdorff dimension of d = 0$d = 0$.
Possibly you wanted to ask: without assuming CH, does a set of positive Hausdorff dimension necessarily have at least continuum cardinality? (I don't know the answer.)
