1. ## countable set

Let $\displaystyle f:[0,1]\rightarrow[0,1]$be such that for every distinct $\displaystyle x_1,x_2,...,x_n$$\displaystyle \in [0,1], \displaystyle f(x_1)+f(x_2)+..+f(x_n)\leq 1. Show that \displaystyle \{x:f(x)>0\} is countable. 2. Originally Posted by Kat-M Let \displaystyle f:[0,1]\rightarrow[0,1]be such that for every distinct \displaystyle x_1,x_2,...,x_n$$\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 $\displaystyle x_i \in E_m$?

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

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

5. 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.

6. Originally Posted by Kat-M
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.
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.

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?
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 $\displaystyle [0,1]$.

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 $\displaystyle E_m$.

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

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