Existence of a set with given Hausdorff dimension Today we had funny discussion whether one can extend the notion of homotopy groups for noninteger indices. Long story short, we asked ourself the following question: > Let $d\in \mathbb{R}_{>0}$. Is it known, whether there exists a set $A\subseteq \mathbb{R}^{\lceil d \rceil}$ such that the Hausdorff dimension of $A$ is equal to $d$? In case this is true, could someone provide me with a reference please.

**First step: answer the question for $0
Take a look at some examples of deterministic fractals. The image below is taken from wiki and quotes this paper, which studies dimensions $1$ and $3$. This is a well-known construction (see fat Cantor set).

![Fat Cantor](

**Second step: from dimension $1$ to $\lceil d \rceil$**

Just use induction and the following result from Mathoverflow: given a separable space $X$ (for instance $X \subset \mathbb{R}^n$ for some $n$), $\mbox{dim}_H(X \times [0,1]) = \mbox{dim}_H(X) + 1$.

Now consider $d > 0$. If $d$ is an integer, just take $A = [0,1]^d$. Otherwise, write $d = \lfloor d \rfloor + \\{d\\}$. Thanks to the first step, as $0<\\{d\\}<1$, we have a Cantor set $C \subset [0,1]$ _s.t._ $\mbox{dim}_H(C) = \\{d\\}$. The previous result yields $\mbox{dim}(C \times [0,1]^{\lfloor d \rfloor}) = \lfloor d \rfloor + \\{d\\} = d$, via induction. And $A := C \times [0,1]^{\lfloor d \rfloor} \subset \mathbb{R}^{\lceil d \rceil}$.