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Intro: The Hausdorff dimension (also known as the Hausdorff–Besicovitch dimension) is an extended non-negative real number associated with any metric space. In general the Hausdorff dimension generalizes the notion of the dimension of a real vector space. That is, the Hausdorff dimension of an n-dimensional inner product space equals $n$. This means, for example, the Hausdorff dimension of a point is zero, the Hausdorff dimension of a line is one, and the Hausdorff dimension of the plane is two. There are, however, many irregular sets that have noninteger Hausdorff dimension. Many of the technical developments used to compute the Hausdorff dimension for highly irregular sets were obtained by Besicovitch.

Intuitively, consider the number $N(r)$ of balls of radius at most $r$ required to cover $X$ completely. When $r$ is small, $N(r)$ is large. For a "well-behaved" set $X$, the Hausdorff dimension is the unique number $d$ such that $N(r)$ grows as 1/rd as r approaches zero. More precise see here>>>.

Having in background the real number line in mind, I am struggling, how to find:

What is the Hausdorff dimension of the set of rational numbers within a certain interval?

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Hint: the rationals are countable. – Anthony Carapetis Aug 27 '13 at 7:24
@AnthonyCarapetis And what about transcendental numbers? Can we calculate H dimension for them? – Филипп Цветков Apr 23 at 21:22
up vote 0 down vote accepted

The answer to the question is provided in the link you give. The Hausdorff dimension of any countable set is $0$.

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thanks got the point. I was confused by the parallels to Cantor set, which is uncountable. – al-Hwarizmi Aug 27 '13 at 13:35

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