This problem has been posted in Mathstack but number of responses is very low (a answer is given but does not look correct).

Consider the following diagram:

Let $0<x<\frac{1}{2}$

Note that $[.]$ is not box function

It is easy to see that $1$ split into $x$ and $(1-x)$. In next stage $x$ split into $x^2$ and $x(1-x)$ and $(1-x)$ split into $x(1-x)$ and $(1-x)^2$. Note that I do not write $x(1-x)$ twice, instead of that in this stage we have three values $x^2,x(1-x),(1-x)^2$. Continue this process in the next stage (see the diagram for more detail)

Now comes the important part:

Add take $1$ first. Then add either $x$ or $(1-x)$. If you added $x$ then next add either $x^2$ or $x(1-x)$ and if you added $(1-x)$ then up next add either $x(1-x)$ or $x^2$. Continue this process again.

For example you will get a value $1+\sum_{n=1}^{\infty}x^n$. Another one could be $1+\sum_{n=1}^{\infty}x(1-x)^{n-1}$. It depends on which path you choose. Every zigzag path will give you a $\color{red}{\text{different sums}}$.

$\color{red}{\text{different sums}}$ means sum over different zigzag path. Note that $\color{red}{\text{different sums}}$ does not mean two different path always give different values.

$\large{F}$or example $1+(1-x)\sum_{n=0}^{\infty}x^n$ and $1+x\sum_{n=0}^{\infty}(1-x)^n$ are two $\color{red}{\text{different sums}}$ but they yeilds same value $2$ i.e. $1+(1-x)\sum_{n=0}^{\infty}x^n=1+x\sum_{n=0}^{\infty}(1-x)^n=2$

##Let say $T_x=\{\text{set of all $\color{red}{\text{different sums}}$ for a given $x$ }\}$

It can be easily seen there is uncountable number of $\color{red}{\text{different sums}}$.

Question: is there uncountable number of distinct elements in $T_x$ for a given $x$?

Observation:

1. It can be easily seen that minimum element of $T_x$ is $\frac{1}{1-x}$ and maximum element of $T_x$ is $\frac{1}{x}$.

2. $2\in T_x$ for every $x\in \Big(0,\frac{1}{2}\Big)$

But I could not find a way to prove or disprove uncountability. Any help would be appreciable.