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$A\subseteq \mathbb{R}^{+}$  is a set of positive real numbers ($0\notin \mathbb{R}^{+}$), for which there exists a positive real number $x$, such that for every finite subset $S\subseteq A$, the sum of the elements in $S$ is less than $x$ .

The question ask to show that $A$ is countable.

I'm not allowed to use calculus in this question.

I looked at $A'=\left \{ \frac{1}{n^{2}}:n\in \mathbb{N} \right \}$

It's known that $\sum_{i=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6}$

I tried to conclude something from $A'$, mainly how to build an injective function from $A$ to $B$ such that $|B|= \aleph_0$ but I didn't make it.

I also consider proof by contradiction, but I didn't make it either.

I would like to know from where to start tackling the problem.

Daniel
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1 Answers1

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Consider the sets $S_n = \{a \in A \: \vert \:a > \frac{1}{n}\}$ for $n \in \mathbb{N}$. $S_n$ must have at most $xn$ elements. (Otherwise, we would be able to pick a finite number of elements of $S_n$ that sum to more than $x$).

Notice that every $a \in A$ must lie in some $S_n$. This means that $A \subseteq \bigcup_{n \in \mathbb{N}} S_n$

Since every $S_n$ has a finite number of elements, it follows that $A$ is a subset of a countable set, hence is countable itself.

Ultra
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  • Thank you very much for the answer, it's beautiful. May I ask how did you came up with $S_n = {a \in A : \vert :a > \frac{1}{n}}$ ? – Daniel May 05 '24 at 19:27
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    @Daniel I came across this idea a few years ago when I was wondering whether the sum of uncountably many positive real numbers is ever finite. https://math.stackexchange.com/questions/20661/the-sum-of-an-uncountable-number-of-positive-numbers – Ultra May 05 '24 at 19:30