Let $\displaystyle f:[0,1]\rightarrow[0,1]$be such that for every distinct $\displaystyle x_1,x_2,...,x_n$$\displaystyle \in [0,1]$, $\displaystyle f(x_1)+f(x_2)+..+f(x_n)\leq 1$.
Show that $\displaystyle \{x:f(x)>0\}$ is countable.
Assume the contrary, i.e. $\displaystyle E = \{x:f(x)>0\}$ is uncountable.
Consider the sets $\displaystyle E_n = \{x: f(x) > 1/n\}$ for $\displaystyle n = 1, 2, 3, \dots$. The union of these sets is $\displaystyle E$, so one of them, say $\displaystyle E_m$, must contain infinitely many points; otherwise $\displaystyle E$ 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 $\displaystyle f(x_1)+f(x_2)+..+f(x_m)$, where $\displaystyle x_i \in E_m$?
The point was to show that there exists an $\displaystyle m\in\mathbb{N}$ such that $\displaystyle E_m := \left\{x\in [0,1]\mid f(x)>\frac{1}{m}\right\}$ is infinite (no matter whether countably infinite or uncountably so). Therefore, by secting finitely many $\displaystyle x_1,\ldots,x_n$ from $\displaystyle E_m$, you can get arbitrarly large sums $\displaystyle f(x_1)+\cdots+f(x_n)$, which contradicts $\displaystyle f(x_1)+\cdots+f(x_n)\leq 1$. Thus, the assumption that $\displaystyle f(x)>0$ for uncountably many $\displaystyle x\in [0,1]$ must be rejected.
even if each$\displaystyle E_m$ is finite, i don know how $\displaystyle f(x_1)+f(x_2)+..f(x_M)\leq 1$could be always true. say$\displaystyle m=3$. If $\displaystyle E_3=\{x:f(x)>1/3\}$ has finitely many elements in it and if you pick 3 elements from this set, then the sum $\displaystyle f(x_1)+f(x_2)+f(x_3)$ is already greater than 1. or i m not supposed to pick points like that? i guess i m not understanding this funtion.
Well, that $\displaystyle f(x_1)+f(x_2)+..f(x_M)\leq 1$ 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.
Well, yes, if it is true that $\displaystyle E_3=\{x\in [0,1]:f(x)>1/3\}$ has at least 3 members, then you get the contradiction $\displaystyle 3\cdot \tfrac{1}{3}=1<f(x_1)+f(x_2)+f(x_3)$ to the assumed property of f that $\displaystyle f(x_1)+f(x_2)+..f(x_n)\leq 1$ for any n different elements $\displaystyle x_1,\ldots,x_n$ of $\displaystyle [0,1]$.say$\displaystyle m=3$. If $\displaystyle E_3=\{x:f(x)>1/3\}$ has finitely many elements in it and if you pick 3 elements from this set, then the sum $\displaystyle f(x_1)+f(x_2)+f(x_3)$ is already greater than 1. or i m not supposed to pick points like that?
Similarly, if we know that $\displaystyle E_m=\{x\in [0,1]:f(x)>1/m\}$, has at least m members (infinitely many, say), then you get the contradiction $\displaystyle 1\leq n\cdot \tfrac{1}{m}<f(x_1)+\cdots f(x_n)\leq 1$, for $\displaystyle n\geq m$ different elements from $\displaystyle E_m$.
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.i guess i m not understanding this function.
But if you want an example of a function that has this property you may consider
$\displaystyle f(x) :=\begin{cases}x,& \text{if }x=\frac{1}{2^n}\text{ for some } n\in\mathbb{N}\backslash\{0\}\\0&\text{otherwise}\ end{cases}$