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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:

Start with $1$. 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.

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:

Start with $1$. 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.

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.

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:

Start with $1$. 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.

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

See this diagram here

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.

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.