Constructing a Function to Prove Countability of a Set of Positive Real Numbers

[moo5003]

Homework Statement

Let A be a set of positive real numbers. Assume that there is a bound b such that the sum of any finite subset of A is less than b. Show that A is countable.

The Attempt at a Solution

So, my first question is what would be a better approach? Trying to find a function from the natural numbers to A or supposing that A is not countable and deriving a contradiction.

I'm a little stumped on how to start the construction of such a function.

 

[Dick]

For all natural numbers n, consider the set of all elements of A that are greater than 1/n, call it A_n. That's a hint. Think about it.

 

[moo5003]

Well, eventually all the A_n's will capture every element of A, its an increasing sequence.

I guess I could argue that each A_n is finite since if it weren't it proabably would be possible to find some finite subset of A_n who'se sum is greater then b.

Since A = the union of A_n's then the cardinlaity of A is countable since its the union of countably many countable sets.

Only problem I have with this argument is I'm not sure how to show with rigor that A_n must be countable or rather (finite).

 

[moo5003]

Thanks, I think I got my proof.

If A_n is infinite then we can find a finite subset of A_n with cardinality m*n where m > b. m,n are natural numbers.

Thus the sum of the subset must be bounded by b. But each element is greater then 1/n thus the sum is greater then (1/n)m*n = m > b which is a contradiction. Thus A_n must be finite.

 

[Dick]

Nice use of the hint.