Set theory problem (uncountable set)

[quasar987]

Homework Statement

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

Homework Equations

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.

 

[Hurkyl]

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

 

[quasar987]

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