Nested family of sets with given Hausdorff dimension Just before (namely here Existence of a set with given Hausdorff dimension) I asked whether one can find for any real number $\alpha>0$ a set $A_\alpha$ such that $A_\alpha$ has Hausdorff dimension $\alpha$. The answer was constructive and used fat Cantor sets and taking the cartesian product with some unit intervall. This made me wonder whether one could push this a bit further, namely

Can we find a family of sets $(A_\alpha)_{\alpha>0}$ such that $A_\alpha$ has Hausdorff dimension $\alpha$, $A_n $ is homotopic to $[0,1]^n$ for $n\in \mathbb{N}$ and $A_\alpha\subset A_\beta$ for $\alpha < \beta$?

 A: Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $A\subseteq \mathbb{R}^n$ and any $0<\alpha < \text{dim}_H(A)$ there exists a compact set $K\subset A$ such that $\alpha= \text{dim}_H(K)$.
From this we get that for all Souslin spaces $A\subset B \subseteq [0;1]$ and any $\gamma \in (\text{dim}_H(A); \text{dim}_H(B)) $ there exists $A\subseteq C \subseteq B$ such that $\text{dim}_H(C)= \gamma$.
Indeed, this follows from the fact that $B\setminus A$ is a Souslin space again and that $\dim_H(B\setminus A)= \dim_H(B)$ as $\dim_H(B) = \dim_H(B\setminus A \cup A)= \max\{ \dim_H(B\setminus A), \dim_H(A) \}$. This allows us to construct (by induction) a family $(A_\alpha)_{\alpha\in [0;1]\cap \mathbb{Q}}$ sucht that $A_\alpha \subset A_\beta$ for $\alpha<\beta$ and $\dim_H(A_\alpha)= \alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $\gamma \in [0;1]$
$$ A_\gamma = \bigcup_{\alpha \in [0;\gamma]\cap \mathbb{Q}} A_\alpha $$
then we get
$$ \dim_H(A_\gamma) = \dim_H\left( \bigcup_{\alpha \in [0;\gamma]\cap \mathbb{Q}} A_\alpha \right) = \sup_{\alpha \in [0;\gamma]\cap \mathbb{Q}} \dim_H(A_\alpha) = \gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]\times \{0 \}$ we play the same game for $\gamma \in [1;2]$, respectively with similar identification for all $\gamma \in [0,\infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
