Hausdorff dimension vs. cardinality What is the relationship between the Hausdorff dimension and cardinality of a set?
Specifically, assuming the Continuum Hypothesis, if a set has Hausdorff dimension greater than zero does, that imply that its cardinality is equal too or greater than that of $2^{\aleph_0}$?
Or, does the negation of CH, imply the existence of a set with positive Hausdorff dimension and cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$?
 A: As stated, countable sets have Hausdorff dimension 0.
So any set $S$ with $\mathrm{HD}(S)>0$ has power $\ge \aleph_1$.
No need for continuum hypothesis.
Without CH, though, we cannot say whether power $ \ge c = 2^{\aleph_0}$ is required.  But this is not about Hausdorff dimension, it is the same question for positive Lebesgue measure in the line.  It is consistent with ZFC (follows from Martin's Axiom) that any set with power $< c$ has Lebesgue measure zero, or (for the same reason, or with the same proof, or even consequently) any set with power $< c$ has Hausdorff dimension zero.  However, without CH (and without Martin's Axiom) there could be sets of reals of power $< c$ but with positive outer Lebesgue measure, and thus Hausdorff dimension 1.
A: The part of the question about the continuum hypothesis (CH) seems confused: without assuming (CH) (but assuming axiom of choice so that cardinals work as they should), $\aleph_1$ is by definition the least uncountable cardinal.  (The continuum hypothesis asserts that $\aleph_1 = 2^{\aleph_0}$.)
Let $X$ be a metric space, and let $x \in X$.  Then it follows immediately from the definition -- see e.g.
http://en.wikipedia.org/wiki/Hausdorff_measure
that for any $d > 0$, the $d$-dimensional Hausdorff measure $H_d(\{x\})$ is equal to zero.  (This is just because a point can be covered by a single ball with arbitrarily small diameter.)  Since $H_d$ is a measure, it is countably additive: thus $H_d(S) = 0$ for any countable set $S$.  If $H_d(S) = 0$, then the Hausdorff dimension of $S$ is at most $d$, so $H_d(S) = 0$ for all 
positive $d$ implies that the Hausdorff dimension of $d = 0$.  
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.)
