# Homework Help: Set theory problem (uncountable set)

1. Oct 3, 2007

### quasar987

1. The problem statement, all variables and given/known data
I would like to show that if we have a non-negative real valued function f defined on f a set X, and that the set of points where f is non-vanishing is uncountable, then for any M > 0, I can find a sequence {x_n} of points in X such that

$$\sum_n f(x_n)>M$$

2. Relevant equations

3. The attempt at a solution
This is not for a set theory class, its in the context of a measure and integration problem where I want to demonstrate a necessary and sufficient condition for a function to be integrable on a set X with respect to the cardinal measure mu(E) = |E|.

Ok, so I would appreciate to get a confirmation that the following assertion is true, because if it is, then I will have solved the problem I think:

"If for any natural number n,

$$f^{-1}(]\frac{1}{n}, +\infty[)$$

is countable, then

$$f^{-1}(]0, +\infty[)$$

is countable too."

It seems so to me since it seems natural that a countable union of countable sets in countable and

$$\bigcup_{n\geq 1} f^{-1}\left(]\frac{1}{n}, +\infty[\right) = f^{-1}\left(\bigcup_{n\geq 1}]\frac{1}{n}, +\infty[\right)=f^{-1}(]0, +\infty[)$$

Thx.

2. Oct 3, 2007

### Hurkyl

Staff Emeritus
A countable union of countables is countable: #(N x N) = #N.

3. Oct 3, 2007

### quasar987

sweet. By the end of the thread I didn't really had a doubt left but thx for confirming, and so fast!