Fat Cantor Set with large complement???

The formula doesn't hold for $a > 1/3$. If you try to remove more than $2^n (1/3)^{n+1}$ at each step $n$, then you will remove the entire interval $[0,1]$ after finitely many steps.

For example, consider $a = 2/5$.

At $n=0$ we remove $2^0 (2/5)^1 = 2/5$. The remaining length is $3/5 = 0.6$.

At $n=1$ we remove $2^1 (2/5)^2 = 0.32$. The remaining length is $0.6 - 0.32 = 0.28$.

At $n=2$ we remove $2^2 (2/5)^3 = 0.256$. The remaining length is $0.28 - 0.256 = 0.024$.

At $n=3$ we remove $2^3 (2/5)^4 = 0.2048$. Oops, no we don't, because that's more than the remaining length.

The set $C_\alpha$ is defined by saying that at the $n$th step, you remove a middle interval of length $\alpha^n$ from each of the remaining intervals. This only makes sense if each of the remaining intervals actually has length at least $\alpha^n$. That will be false for sufficiently large $n$ for any $\alpha>1/3$, and so $C_\alpha$ is only defined for $\alpha\leq 1/3$.

For an explicit example, if $\alpha=0.4$, then we first remove an interval of length $0.4$ to leave $[0,0.3]$ and $[0.7,1]$. For subsequent steps, let's keep track of just the leftmost of our intervals. We next remove intervals of length $0.16$ to leave $[0,0.07]$ as our leftmost interval. Then we remove intervals of length $0.064$ to leave $[0,0.003]$ as our leftmost interval. Next we need to remove intervals of length $0.0256$, but we can't, because our remaining intervals have length $0.003$!

Note that in fact you can see that the intervals will always be long enough iff the sum $ \sum_{n=0}^\infty \alpha(2\alpha)^n$ is at most $1$, since each new term we add represents the total length of intervals we need to remove at the next step, and there is enough length left iff the resulting partial sum is still at most $1$. So that's where the bound $\alpha\leq 1/3$ comes from: when $\alpha=1/3$, the sum converges to $1$, and any larger $\alpha$ makes it greater than $1$ so the intervals eventually become too small.