As everyone knows, the standard middle-thirds Cantor set is constructed by dividing the interval into three equal parts, removing the middle one, then applying the same procedure to the remaining two intervals, etc. 


The resulting set has Hausdorff dimension $s=\log 2/\log 3$, in view of Hutchinson's formula:
$$
\sum\_{i=1}^m r\_i^s=1,
$$
where in our case $m=2, r\_1=r\_2=1/3$. 


Now, assume that $m=2$ but that on level $n$ we remove the middle-thirds interval whose relative measure is not exactly $1/3$ but $1/3+\delta\_n$, where $\delta\_n\to0$ sufficiently fast. (It may also depend on a position of the interval but there is always a uniform upper bound which tends to 0 sufficiently fast.) 


Is it still true that the Hausdorff dimension of such a set is $\log2/\log3$? 


This is actually a `toy question', since in my set-up the corresponding iterated function system is infinite countable. However, Hutchinson's formula works for such IFS just as well (with $m=\infty$), so I'm sure the conclusion should be the same. If it helps, the uniform upper bound for the $\delta\_n$ in my case is a double exponent, i.e., $\frac1n \log (-\log \delta\_n)\to \text{const}$ as $n\to+\infty$. 



Hutchinson's formula for asymptotically homogeneous Cantor sets
Since you remove "more", it should be more or less clear that $\text{HD}\; C\_\delta \le \text{HD}\; C$ (where $C$ is the standard Cantor set and $C\_\delta$ is the "perturbation" you describe). For proving the opposite inequality it is enough to exhibit a measure on $C\_\delta$ whose Hausdorff dimension is equal to $\text{HD}\; C$. This is the usual uniform measure $m$ on $C\_\delta$ (i.e., the one whose value is $1/2^n$ on each of the rank $n$ intervals), because if $\delta\_n\to 0$ fast enough, then 
$$
\frac{\log m B(x,r)}{\log r} \to \log 2/\log 3
$$ 
for $m$-a.e. $x\in C\_\delta$.


