Density of a particular set of numbers Let us consider
$$Z = X_1 + X_1 X_2 + X_1 X_2 X_3+\cdots$$
where $X_i \in \{\alpha, \beta\}$ with $0 < \alpha < \beta <1$. In short, the $X_i$'s are arbitrary but can only take on two distinct values: $\alpha$ or $\beta$. This system is described in details here. The average value of $Z$, computed over all the infinitely many combinations of the $X_i$'s, is $\mu/(1-\mu)$ with $\mu = (\alpha + \beta)/2$. In most cases, the distribution for $Z$ (its support domain) is not a dense set: try for instance with $\alpha =0, \beta = 0.5$. However, when $\beta - \alpha$ is rather small (say $\alpha=0.70,\beta =0.72$), the support domain of $Z$ seems to be dense in $[\min(Z), \max (Z)]$. 
Main Question
Can the support domain of $Z$ sometimes be dense, as suggested? What are the conditions on $\alpha, \beta$ for this to happen?
Auxiliary Question
Assume the $X_i$'s are i.i.d random variables, taking the value $\alpha$ or $\beta$ with the same probability $\frac{1}{2}$. What is the distribution of
$$Z_\alpha = \lim_{\beta\rightarrow\alpha} \frac{Z-\min(Z)}{\beta-\alpha}$$
This limit may not exist, but if $\alpha=\frac{1}{2}$ it looks like the limiting distribution might be uniform. If it is indeed uniform, it would be uniform on $[0, 4]$.
 A: I will split my answer into two parts: the main question, and the auxiliary question.
Main question
The formula $Z = X_1+ X_1 X_2 + X_1 X_2 X_3 + \cdots$ corresponds to a particular numeration system, described in a question that I posted today, here. In short, if $\alpha + \beta = 1$, the distribution of $Z$, assuming $P(X_i = \alpha) = \alpha$ and $P(X_i = \beta) = \beta$, is uniform on $[\frac{\alpha}{1-\alpha},\frac{\beta}{1-\beta}]$. Any real number in that interval has a unique representation as $Z$ in that system. Thus, the support domain is dense in that interval if $\alpha + \beta =1$. If $\alpha + \beta < 1$, it is not dense. If $\alpha + \beta \geq 1$, it should be dense.  
Auxiliary question 
First note that if $Z$ is uniform, obviously it would be on $[\frac{\alpha}{1-\alpha},\frac{\beta}{1-\beta}]$. Then we would have
$$E(Z) = \frac{1}{2} .\Big(\frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta} \Big) = \frac{\alpha + \beta - 2\alpha\beta}{2(1-\alpha)(1-\beta)}\mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } (\star)$$
We also have, assuming $P(X=\alpha) = 1-p$ and $P(X=\beta) = p$: 
$$E(Z) = \frac{E(X)}{1- E(X)} = \frac{(1-p)\alpha+p\beta}{1-(1-p)\alpha -p\beta } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } \mbox{ } (\star\star)$$
Here $X$ represents any of the $X_i$'s. Likewise, we have a similar pair of equations for the variance, see here. 
The only way we can have $(\star) = (\star\star)$ both for the expectation, variance and third moment is if $p=\beta$ and $\alpha+\beta = 1$. And in that case, it turns out (as pointed out in my answer to the main question) that $Z$ is indeed uniform. 
Now for the limit as $\beta \rightarrow \alpha$, let's proceed as follows. Let us use the notation $\beta = \alpha + \epsilon$ with $\epsilon \rightarrow 0$. We have
$$Var(Z) = \frac{Var(X)}{(1-E(X^2))(1-E(X))^2}.$$
Again, see here (section 1) for this result. It is easy to show that 
$$Var(X) = p(1-p)\epsilon^2$$
regardless of $p$. As $\beta\rightarrow \alpha$, we thus have:
$$Var\Big(\frac{Z - \min(Z)}{\beta-\alpha}\Big) \rightarrow \frac{p(1-p)}{(1-\alpha^2)(1-\alpha)^2} < \infty$$
$$E\Big(\frac{Z - \min(Z)}{\beta-\alpha}\Big)\rightarrow \frac{p}{(1-\alpha)^2}$$
This indicates that the limiting distribution exists, regardless of the values of $p$ and $\alpha$, but it does not prove that if $p=\alpha =\frac{1}{2}$, that distribution is uniform. Based on the variance, if $Z^* = (Z-\min(Z))/(\beta - \alpha)$ is uniform on $[0, \lambda]$ as $\beta \rightarrow\alpha$, then we must have:
$$\lambda = \sqrt{12 Var(Z^*)}=\frac{\sqrt{12 p (1-p)}}{(1-\alpha)\sqrt{1-\alpha^2}} .$$
Based on the expectation, we must also have, as $\beta \rightarrow\alpha$:
$$\lambda = 2E(Z^*)=\frac{2p}{(1-\alpha)^2}.$$
The only way the two values of $\lambda$ are identical is if
$$\alpha = \frac{3-4p}{3-2p},$$
which also implies $p< 0.75$. We also discovered (see answer to main question) that if $\alpha + \beta \neq 1$, then $Z$ may not be uniform, though this does not say anything about the limit $Z^*$ as $\beta \rightarrow\alpha$. Assuming this is also true for the limit, it means at the limit, $\alpha = \beta =\frac{1}{2}$ to satisfy the requirement $\alpha + \beta = 1$. This in turn implies that $p=\frac{1}{2}$ (based on the above formula) and thus $\lambda = 4$.
Looking at higher moments may completely settle this problem. The third moment alone will most likely settle this discussion. It involves cumbersome but elementary, high-school level computations. Note that $E(Z^3)$ can be computed using the formula $E(Z^3) = E(X^3)E[(1+Z)^3]$.
