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Let f be a real-valued function defined for every $x$ in the interval $0\le x \le 1$. Suppose there is a positive number M having the following property: for every choice of a finite number of points $x_1, x_2, ..., x_n$ in the interval $0 \le x \le 1$, the sum $$|f (x_1) + ... + f (x_n)| \le M$$ Let $S$ be the set of those $x$ in $0 \le x \le 1$ for which $f(x) \neq 0$. Prove that $S$ is countable.

I am having hard time to understand it's solution. The solution is as follows.

Proof : Let $S_n = \{x \in [0, 1] : |f (x)| \ge 1/n\}$ , then $S_n$ is a finite set by hypothesis. In addition, $S = \cup_{n=1}^\infty S_n$. So, S is countable.

There are infinite irrationals between $0$ and $1$, how does defining taking $x$ such that $|f(x)| \ge 1/n$ prove that the set is countable. Aren't we counting irrationals as well as rationals between $0$ and $1$?

I think it would be more intuitive to begin by assuming that the set $S$ in uncountable and arrive at contradiction that $|f (x_1) + ... + f (x_n)|$ is bounded. But I don't know how. Can we do this way?

Also can anyone elaborate that proof from manual so that I can understand?

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1  
$\{x : |f(x)|> 0\} =\cup_{n=1}^\infty S_n$. –  David Mitra Jan 19 '13 at 12:43
    
how do i show that? –  Santosh Linkha Jan 19 '13 at 12:44
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If $|f(x)|=b>0$, then there is an $n$ with $1/n<b$. Then $|f(x)|>1/n$, so $x\in S_n$. So $\{x: |f(x)|>0\}\subset\cup_{n=1}^\infty S_n $. The other containment is easy (if $x\in \cup_{n=1}^\infty S_n$, then $x\in S_m$ for some $m$ ...). –  David Mitra Jan 19 '13 at 12:50

1 Answer 1

up vote 2 down vote accepted

To see that each $S_n$ is finite, suppose it is at least countable. Then there are either at least countably many positive or negative $x$ in it, but without loss of generality assume there are positive. Then pick an integer $k$ such that $k/n > M \Rightarrow k > Mn$. Then since every $f(x) > 1/n$, and by hypothesis there are infinitely many such $x$ in $S_n$, pick $k$ of them, and then $|f(x_1) + ... + f(x_{k})| > | 1/n + ... \text{k times} ... + 1/n | > k/n > M$ contradicting the hypothesis, that is, showing that there are no more than $k$ positive numbers in $S_n$.

That method does count both irrationals and rationals. For any $x$ such that $|f(x)|>0$, it is contained in $S_n$ as soon as $|f(x)| > 1/n$, which since $\{1/n\}_{n\in\mathbb{N}}$ converges to $0$ is bound to happen. The proof has nothing specific to do with rationals, it just picked a convenient positive sequence which converges to $0$, and any other would just as well worked (say $\{\pi / n\}_{n\in\mathbb{N}}$).
The proof the proceeds by noting that union of countable collection of finite sets is countable. You can see a proof of a stronger fact here.

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i see, so it uses the fact that the bounded implies a sum (or sums) of converging sequence(s)?? –  Santosh Linkha Jan 19 '13 at 12:46
    
Sorry, I'm not sure what part of the proof you are referring to. Are you having trouble understanding why each $S_n$ is finite? –  user5501 Jan 19 '13 at 12:49
    
yes!! I am having problem on that –  Santosh Linkha Jan 19 '13 at 12:51
    
@experimentX: I have updated my answer to address those concerns. –  user5501 Jan 19 '13 at 12:59
    
thanks!! that was understandable!! –  Santosh Linkha Jan 19 '13 at 13:01

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