From Covers to Gales
In this post we attempt to understand a relationship between Hausdorff dimension defined in a classical way and Hausdorff dimension defined using gales. Recall that classical Hausdorff dimension is defined in terms of covers. So, in order to understant that relationship, we need to be able to 'switch' back and forth from covers and gales. In this post we will try to switch from covers to gales.
As before, we work in Cantor space $\mathcal{C} = \{0,1\}^{\mathbb{N}}$. We start with some subset $X\subseteq \mathcal{C}$. For any given $\varepsilon>0$, one can choose $s\ge 0$ such that $s-\varepsilon < \dim_H(X) < s$. Then we go on to show that there $s$-gale succeeding on $X$, i.e. $\inf G(X)\le s$. Combining this with above inequality, we have
\[
\inf G(X)\le s < \dim_H(X) + \varepsilon
\]
By letting $\varepsilon \to 0$, one shows that $\inf G(X)\le \dim_H(X)$. So, we are left to show that it is indeed possible to construct $s$-gale given the fact that $\dim_H(X) < s$. Let us proceed with this construction.
Construction
Instead of a single $s$-gale, we are going to construct an infinite family of them. At first, it might seem counter-intuitive, however, the construction process will reveal the idea behind it. Having an infinite family of $s$-gales $\{d_i\}_{i=1}^{\infty}$, we are going to consolidate them into one $s$-gale by setting:
\[
d = \sum_{i=1}^{\infty} 2^{-i}d_i
\]
The output of above linear combination is going to be $s$-gale again, simply because of the fact that $s$-gales are closed under linear addition and scalar multiplciation.
Construction of individual gales
To construct each $d_i$ we consider a cover $C_i = \{C_{w_j}\}_{j=1}^{\infty}$ such that
\[
\sum_{j=1}^{\infty} [diam(C_{w_j})]^s \le 2^{-2i}
\]
Observe that above cover exists because $\dim_H(X)<s$. From above condition, it can inferred that $2^{2i}\sum_{i=1}^{\infty} 2^{-\vert w_j \vert s} \le 1$. So we construct an $s$-gale $d_i$ with the norm $2^{2i}\sum_{i=1}^{\infty} 2^{-\vert w_j \vert s}$, which is pre-allocates $2^{2i}2^{-\vert w_j \vert s}$ for the string $w_j$. The sum allocated for each $w_j$, then, gets multiplied by the factor of $2^s$ at each level until it reaches $w_j$. This way we can ensure that $d(w_j) = 2^{2i}$. As for strings that do not appear in $C_i$ or those which extend $w_j$, we do not care much. Also one could note that this procedure might require the covering $C_i$ to be prefix-free. However, it is something we can alsways assume.
Consolidation
Having constructed all $d_i$'s we can now start consolidating them into one $s$-gale, $d$. As we have mentioned earlier $d = \sum_{i=1}^{\infty} 2^i d_i$. To start with it is worth observing that
To show that $d$ indeed succeeds on $X$, let us pick some $\xi\in X$, we need to show that $d$ succeeds on $\xi$, i.e.
\[
\limsup_n d(\xi[n]) = \infty
\]
Since each cover $C_i$ mentioned earlier covers $X$, we have that there is some $w_j\in C_i$ such that $w_j\prec \xi$. In turn this means that $d(\xi|_{w_j})\ge 2^i$. Since last argument is true for all $i\in \mathbb{N}$, we indeed have that $d$ succeeds on $\xi$.
This completes the proof. In this way or another, one can switch from covers to gales. In the next posts we hope to share how can one switch from gales to covers.
As before, we work in Cantor space $\mathcal{C} = \{0,1\}^{\mathbb{N}}$. We start with some subset $X\subseteq \mathcal{C}$. For any given $\varepsilon>0$, one can choose $s\ge 0$ such that $s-\varepsilon < \dim_H(X) < s$. Then we go on to show that there $s$-gale succeeding on $X$, i.e. $\inf G(X)\le s$. Combining this with above inequality, we have
\[
\inf G(X)\le s < \dim_H(X) + \varepsilon
\]
By letting $\varepsilon \to 0$, one shows that $\inf G(X)\le \dim_H(X)$. So, we are left to show that it is indeed possible to construct $s$-gale given the fact that $\dim_H(X) < s$. Let us proceed with this construction.
Construction
Instead of a single $s$-gale, we are going to construct an infinite family of them. At first, it might seem counter-intuitive, however, the construction process will reveal the idea behind it. Having an infinite family of $s$-gales $\{d_i\}_{i=1}^{\infty}$, we are going to consolidate them into one $s$-gale by setting:
\[
d = \sum_{i=1}^{\infty} 2^{-i}d_i
\]
The output of above linear combination is going to be $s$-gale again, simply because of the fact that $s$-gales are closed under linear addition and scalar multiplciation.
Construction of individual gales
To construct each $d_i$ we consider a cover $C_i = \{C_{w_j}\}_{j=1}^{\infty}$ such that
\[
\sum_{j=1}^{\infty} [diam(C_{w_j})]^s \le 2^{-2i}
\]
Observe that above cover exists because $\dim_H(X)<s$. From above condition, it can inferred that $2^{2i}\sum_{i=1}^{\infty} 2^{-\vert w_j \vert s} \le 1$. So we construct an $s$-gale $d_i$ with the norm $2^{2i}\sum_{i=1}^{\infty} 2^{-\vert w_j \vert s}$, which is pre-allocates $2^{2i}2^{-\vert w_j \vert s}$ for the string $w_j$. The sum allocated for each $w_j$, then, gets multiplied by the factor of $2^s$ at each level until it reaches $w_j$. This way we can ensure that $d(w_j) = 2^{2i}$. As for strings that do not appear in $C_i$ or those which extend $w_j$, we do not care much. Also one could note that this procedure might require the covering $C_i$ to be prefix-free. However, it is something we can alsways assume.
Consolidation
Having constructed all $d_i$'s we can now start consolidating them into one $s$-gale, $d$. As we have mentioned earlier $d = \sum_{i=1}^{\infty} 2^i d_i$. To start with it is worth observing that
- $d(\varepsilon) \le 1$;
- $d(w_j)\le 2^i$ for $w_j$ which appears in $C_i$.
To show that $d$ indeed succeeds on $X$, let us pick some $\xi\in X$, we need to show that $d$ succeeds on $\xi$, i.e.
\[
\limsup_n d(\xi[n]) = \infty
\]
Since each cover $C_i$ mentioned earlier covers $X$, we have that there is some $w_j\in C_i$ such that $w_j\prec \xi$. In turn this means that $d(\xi|_{w_j})\ge 2^i$. Since last argument is true for all $i\in \mathbb{N}$, we indeed have that $d$ succeeds on $\xi$.
This completes the proof. In this way or another, one can switch from covers to gales. In the next posts we hope to share how can one switch from gales to covers.
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