1. ## countable set

Let $f:[0,1]\rightarrow[0,1]$be such that for every distinct $x_1,x_2,...,x_n$ $\in [0,1]$, $f(x_1)+f(x_2)+..+f(x_n)\leq 1$.
Show that $\{x:f(x)>0\}$ is countable.

2. Originally Posted by Kat-M
Let $f:[0,1]\rightarrow[0,1]$be such that for every distinct $x_1,x_2,...,x_n$ $\in [0,1]$, $f(x_1)+f(x_2)+..+f(x_n)\leq 1$.
Show that $\{x:f(x)>0\}$ is countable.
Assume the contrary, i.e. $E = \{x:f(x)>0\}$ is uncountable.

Consider the sets $E_n = \{x: f(x) > 1/n\}$ for $n = 1, 2, 3, \dots$. The union of these sets is $E$, so one of them, say $E_m$, must contain infinitely many points; otherwise $E$ could be written as the union of countably many finite sets and would be countable, contrary to our assumption.

Now what does that say about the sum $f(x_1)+f(x_2)+..+f(x_m)$, where $x_i \in E_m$?

3. i understand the union of $E_m$ make up the whole set $E$ but i m still not getting why each $E_m$ is countable. how does $f(x_1)+f(x_2)+...+f(x_n)\leq1$ help me figure it out?

4. Originally Posted by Kat-M
i understand the union of $E_m$ make up the whole set $E$ but i m still not getting why each $E_m$ is countable. how does $f(x_1)+f(x_2)+...+f(x_n)\leq1$ help me figure it out?
The point was to show that there exists an $m\in\mathbb{N}$ such that $E_m := \left\{x\in [0,1]\mid f(x)>\frac{1}{m}\right\}$ is infinite (no matter whether countably infinite or uncountably so). Therefore, by secting finitely many $x_1,\ldots,x_n$ from $E_m$, you can get arbitrarly large sums $f(x_1)+\cdots+f(x_n)$, which contradicts $f(x_1)+\cdots+f(x_n)\leq 1$. Thus, the assumption that $f(x)>0$ for uncountably many $x\in [0,1]$ must be rejected.

5. even if each $E_m$ is finite, i don know how $f(x_1)+f(x_2)+..f(x_M)\leq 1$could be always true. say $m=3$. If $E_3=\{x:f(x)>1/3\}$ has finitely many elements in it and if you pick 3 elements from this set, then the sum $f(x_1)+f(x_2)+f(x_3)$ is already greater than 1. or i m not supposed to pick points like that? i guess i m not understanding this funtion.

6. Originally Posted by Kat-M
even if each $E_m$ is finite, i don know how $f(x_1)+f(x_2)+..f(x_M)\leq 1$could be always true.
Well, that $f(x_1)+f(x_2)+..f(x_M)\leq 1$ is a given fact about the particular function f in question. There is no need to prove that it holds. You just assume it holds and show, what you are asked to show.

say $m=3$. If $E_3=\{x:f(x)>1/3\}$ has finitely many elements in it and if you pick 3 elements from this set, then the sum $f(x_1)+f(x_2)+f(x_3)$ is already greater than 1. or i m not supposed to pick points like that?
Well, yes, if it is true that $E_3=\{x\in [0,1]:f(x)>1/3\}$ has at least 3 members, then you get the contradiction $3\cdot \tfrac{1}{3}=1 to the assumed property of f that $f(x_1)+f(x_2)+..f(x_n)\leq 1$ for any n different elements $x_1,\ldots,x_n$ of $[0,1]$.

Similarly, if we know that $E_m=\{x\in [0,1]:f(x)>1/m\}$, has at least m members (infinitely many, say), then you get the contradiction $1\leq n\cdot \tfrac{1}{m}, for $n\geq m$ different elements from $E_m$.

i guess i m not understanding this function.
You need not understand this function: you only have to assume that it has the properties that you are told it has and move on to show what you are told to show.

But if you want an example of a function that has this property you may consider

$f(x) :=\begin{cases}x,& \text{if }x=\frac{1}{2^n}\text{ for some } n\in\mathbb{N}\backslash\{0\}\\0&\text{otherwise}\ end{cases}$