Does interval spacing effect Hausdorff dimension of Cantor set? Let $C=\bigcap_{j=0}^{2^n}C_j$, $C_0=[0,1]$, and the intervals in the construction of each stage of $C_j$ consists of removing the center 1/3 from the $j-1$ stage intervals.  In other words, the Cantor middle third set.  The dimension of this set is known to be $\frac{\log 2}{\log 3}$.  Now consider the following adjustment: we now only require allow the spacing between the intervals at each stage, $j$, to be disjoint. That is, we only require spacing between intervals to be >0. It is clear to me, from the definition of Hausdorff dimension that interval spacing should not affect an upper bound calculation on a Cantor Set. 
My question is this: Can changing the nonnegative spacing between intervals cause the lower bound calculation to diverge from the upper bound calculation in the context of Cantor sets? For sets in general? If yes for the Cantor sets, can someone direct me to a lower bound technique that explicitly incorporates spacing. If yes only for sets in general, what about Cantor sets makes spacing irrelavant?
 A: I regret having to note that the other answer is wrong. It is well known, since work of Moran in 1946, that the Hausdorff dimension of the particular set that you describe does not depend on the location of the intervals at each step. For a proof you can see the book Dimension and Recurrence in Hyperbolic Dynamics, by Barreira (see Theorem 3.1.1).
A: Update: see the other answer

You will not be able to get the same lower bound, and for a good reason. Reduced spacing between the remaining 1/3-intervals can make the Hausdorff dimension of the Cantor-type set arbitrarily small. Intuitively, this is because when the intervals are placed nearby, they can be covered more efficiently. Here is a more precise argument. 
Fix $s>0$; I will build a Cantor-type set of $s$-dimensional Hausdorff measure zero. To do this, I will arrange to  have a sequence of scales $\delta_j\to 0$ such that generation $m_j$ of the construction can be covered by $N_j$ intervals of size $\delta_j$, with $N_j \delta_j^s\le 1/j$. 
For $j=1$ I can take $\delta_1=N_1=1$, which cover the $0$th generation $(m_j=0)$ of the Cantor-type set, namely $[0,1]$. Suppose now that $N_j$, $\delta_j$, $m_j$ are already arranged. 
I have $2^{m_j}$ intervals in the $m_j$-generation. Let $N_{j+1}=2^{m_j}$ and choose $\delta_{j+1}$ so that  $N_{j+1} \delta_{j+1}^s\le 1/(j+1)$. 
For each of the aforementioned $2^{m_j}$ intervals, repeat the process of replacing the interval by two intervals of $1/3$ its length and leaving only a tiny gap between them: after sufficiently many such steps the resulting set will be contained in $\frac12\delta_{j+1}$-neighborhood of the midpoint of the original interval. At this moment I have generation $m_{j+1}$ of the Cantor-type set, which is covered by $N_{j+1}$ intervals of length $\delta_{j+1}$, as desired.
