Prove that there is $\{x_n\}_{n=1}^{\infty}$ such that $ \sum_{n=1}^{\infty}f(x_n)=\infty$

Question

I have to prove the following statement, but I am not sure that my solutions is correct:
Let $f:[0,1] \rightarrow\mathbb{R}$ be a function such that $f(x)>0$ for all $x\in[0,1]$, but which is otherwise arbitrary. Prove that there is a sequence $\{x_n\}_{n=1}^{\infty}$ of distinct elements ${x_n}\in [0,1]$ such that$$ \sum_{n=1}^{\infty}f(x_n)=\infty$$

My solution goes as follows: let $A_n=\{x :f(x)>1/n\}$. Notice that for some $n_{\ast}\in\mathbb{N} $ we have $A_n\not=\varnothing$ for every $n\geq n_\ast$. Moreover, $A_n\subseteq A_{n+1}$ which implies $\bigcup_{i=1}^{n}A_i=A_n$. Since by hypothesis $f(x)>0$ for every $x\in [0,1]$ it is easy to prove that $$\bigcup_{n\in \mathbb{N}}A_n=[0,1]$$ But this implies that for some $k \in \mathbb{N}$ the set $\bigcup_{i=1}^{k}A_i=A_{k}$ must be uncountable, otherwise $[0,1] $ would be the countable union of countable sets and therefore a countable set. But since every infinite set contains a countable (infinite) set, there is a sequence $\{x_m\}_{m\in \mathbb{N}}$ such that $x_m\in A_k$ for every $m\in \mathbb{N}$ and therefore $f(x_m)>1/k $. Hence:$$\sum_{m=1}^\infty f(x_{m})\geq \sum_{n=1}^{\infty}1/k= \infty$$

Ops, I forgot. The elements of the sequence must be distinct. Thanks for the comment! — Lorenzo Stanca, Mar 23 '14 at 21:23
Your solution is correct. Do you really need the fact $\cup_{i=1}^n A_i = A_n$ (and the same question for the fact about nonemptiness)? — Antoine, Mar 23 '14 at 21:38
@Antoine About the nonemptiness, I think it is necessary although a bit trivial. I just mentioned it because previously I thought I had found a solution that used the axiom of choice, but I realized it wasn't correct. Notice that the axiom choice is still being used to prove this theorem since it is necessary to prove that any infinite set has a countable subset. — Lorenzo Stanca, Mar 23 '14 at 21:49

score 1 · Accepted Answer · answered Mar 23 '14 at 21:50

You're solution is correct. As others have said, you could have just jumped from defining $A_n$ to immediately noting that $[0,1]=\cup A_i$. Another way of restating the main idea of your proof is that it's impossible for all but finitely many numbers be sent to less than $\frac{1}{n}$ for every $n$.

Prove that there is $\{x_n\}_{n=1}^{\infty}$ such that $ \sum_{n=1}^{\infty}f(x_n)=\infty$

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