Showing that the set $S$ is countable Here's the question:
Let $S$ be a set of positive real numbers. There is a constant $M>0$ such that for any $\{ a_1,a_2,\ldots, a_k \}\subset S$, we have that
$$\sum_{n=1}^{k}a_n\le M \tag{1}$$
Prove that $S$ is at most countable.

Here's my attempt:
If $S$ is finite then there is nothing to prove. So, we prove the following:
Claim. If $S$ is any infinite set of positive real numbers satisfying condition $(1)$ then $\max S$ exists.
Proof. Let $S$ be a set satisfying $(1)$. Then $M$ is an upperbound for the set $S$. By least upper bound axiom, $\alpha=\sup S$ exists. Since $S$ is a set of positive reals, $\alpha >0$. If the maximum existed then it must be equal to $\alpha$. So, suppose that it does not exist. Choose an $\varepsilon >0$ such that $a-\varepsilon >0$. Then by the Archimedean property, $M\le k(\alpha - \varepsilon)$ for some $k\in \mathbb{N}$.
Now, since $\alpha$ is the supremum and not the maximum, we can select distinct elements $a_1,a_2, \ldots , a_k$ from $S$ such that $\alpha - \varepsilon < a_i < \alpha$ for each $i$. Then $\sum_{n=1}^{k}a_n> k(\alpha - \varepsilon ) \ge M$ but this would contradict our assumption that $(1)$ holds. So, $\alpha$ must be the maximum. $\blacksquare$
Now, let $S$ be any set of positive real numbers satisyfing $(1)$. Then let $a_1= \max S$. Also $S \setminus \{ a_1 \}$ satisfies $(1)$ so, let $a_2=\max (S\setminus\{ a_1 \})$. We repeat this argument to obtain a sequence $\{ a_k \}$ such that $a_k=\max (S\setminus \{ a_1, a_2 , \ldots, a_{k-1} \})$.
Clearly this $\{ a_k \}$ is decreasing and there is no $a\in S$ such that $a_{k+1}<a<a_k$ for any $k$. Also, $\sum_{k=1}^{\infty}a_k$ converges, so, $a_n \to 0$.
If $a \in S\setminus \{a_k\}_{k\in \mathbb{N}}$ then $a<a_k$ for each $k$. So $a\le0$ (by taking limits) which is not possible. So $S=\{ a_k\}_{k\in \mathbb{N}}$.

Is this proof correct? Alternative proofs are also welcome!
 A: It wouldn’t hurt to say something to justify the assertion that $\sum_{n\ge 1}a_n$ converges, but the proof is correct. (There is a typo in that sum though: you have $\sum_{\color{red}k=1}^\infty a_{\color{red}n}$.) Here’s a slightly different approach.
Let $\alpha=\sup S>0$. For $n\in\Bbb N$ let $I_n=\left(2^{-(n+1)}\alpha,2^{-n}\alpha\right]$; then
$$S\subseteq(0,\alpha]=\bigcup_{n\ge 0}I_n\,,$$
and the intervals $I_n$ are pairwise disjoint. There are only countably many of these intervals, so if $S$ is uncountable, there must be an $n\in\Bbb N$ such that $S\cap I_n$ is uncountable. But then if $K$ is a $k$-element subset of $A\cap I_n$, we have
$$\sum K>2^{-(n+1)}\alpha k\,,$$
and this can be made arbitrarily large by choosing $k$ large enough, contradicting $(1)$.
A: It looks good to me, an alternative proof could be as follow. Suppose $S$ is not countable and let for $n\geqslant 1$, $S_n=\left\{s\in S,s\geqslant\frac{1}{n}\right\}$. Clearly $\displaystyle S=\bigcup_{n\geqslant 1}S_n$ and there is at least one $p\geqslant 1$ such that $S_p$ is not countable, otherwise $S$ would be countable as a countable union of countable sets. In particular $S_p$ is infinite so we can take $\{s_1,\ldots,s_k\}\subset S_p\subset S$ with $k>pM$. We thus have $\displaystyle\sum_{n=1}^k s_n\geqslant\frac{k}{p}>M$ which is not possible and therefore $S$ is countable.
