How many sub-sums of the harmonic series converge to a given $x$?

Question

Pick some positive $x$. $\hspace{0.5mm}$ Let $\mathcal{C}_x$ be the set of all subsets $S \subset \mathbb{Z}^{+}$ such that $\displaystyle{\hspace{1mm} \sum_{n \in S} \frac{1}{n} = x}$.

Is $\mathcal{C}_x$ countable?

(There are follow-ups and generalizations of this question one might ask, too, which I omit here to keep it neat. $\hspace{0.5mm}$ However, answers to more general versions of this question are welcome.)

Answer 1

The set $\mathcal C_x$ must be uncountable.

For any $x>0$ and $S\in\mathcal C_x,$ we set $S_*=\{1/n;~n\in S\}$ and $$I_x=\{S_*;~\hbox{$S$ is an infinite set}\}.$$ Then, $I_x\neq\emptyset,$ and $I_x$ is indeed an infinite set. For convenience, we may assume that each element of $I_x$ is a decreasing sequence. We are going to show that $I_x$ is a perfect set in the Banach space $l^1$.

• $I_x\subset l^1$ is a closed subset. For any $n\in\mathbf N_+$ and $A_n=(a_{n,1},a_{n,2},\ldots)\in I_x$ with $$\sum_{k\geq1}a_{n,k}=x,$$ we suppose that $\{A_n\}$ converges to $a=(a_1,a_2,\ldots)$ in $l^1.$ Then, $$\sum_{k\geq1}|a_{n,k}-a_k|\to0\quad\hbox{as $n\to\infty,$}$$ which implies that $$\sum_{k\geq1}a_k=x,$$ so we suffice to show that $a_k>0$ for every $k\in\mathbf N_+.$ By convention, we deduce that $a$ is a decreasing sequence. Suppose $j$ is the least index such that $a_j=0.$ Then, $a_i>0$ for any $1\leq i\leq j-1$, and $a_i=0$ for any $i\geq j.$ Since $$a_i=\lim_{n\to\infty}a_{n,i},$$ and we note that $$a_{n,i}\in\{1,1/2,1/3,\ldots\},$$ and $0$ is the unique limit point of $\{1,1/2,1/3,\ldots\}.$ We then obtain that $a_i=a_{n,i}$ for $1\leq i\leq j-1$ and some large $n.$ Thus, $$\sum_{1\leq i\leq j-1}a_{n,i}=x,$$ which is a contradiction as $A_n$ is an infinite set. By the same reason, we get that $a$ is a strictly decreasing sequence. Hence, $a\in I_x$.
• $I_x$ has no isolated point, which is because that each element of $I_x$ is an infinite set. The proof is not difficult but lengthy, and so we omit the details.

Answer 2

Let $(a_n)_{n\in\Bbb N}$ be a sequence of non-negative reals such that $\lim_{n\to\infty}a_n=0$ and $\sum_{n\in\Bbb N}a_n=\infty.$

Let $x>0.$

We can find disjoint $A_1,B_1\subset \Bbb N$ such that $\sum_{j\in A_1}a_j=x/2=\sum_{j\in B_1}a_j.$

For each $n\in\Bbb N$ we can find disjoint $A_{n+1}, B_{n+1}\subset\Bbb N$ such that $A_{n+1}\cap (\cup_{j\le n}(A_j\cup B_j))=\emptyset =B_{n+1}\cap (\cup_{j\le n}(A_j\cup B_j))$ and such that $\sum_{j\in A_{n+1}}a_j=x2^{-n-1}=\sum_{j\in B_{n+1}}a_j.$

Now for $ f:\Bbb N\to\{0,1\}$ and $n\in\Bbb N$ let $f^*(n)=A_n$ if $f(n)=0,$ or $f^*(n)=B_n$ if $f(n)=1.$ And let $\psi (f)= \cup_{n\in\Bbb N}f^*(n).$ Then $$\sum_{n\in\psi(f)}a_n=x.$$

If $f:\Bbb N\to\{0,1\}$ and $g:\Bbb N\to\{0,1\}$ with $f\ne g$ then $\psi(f)\ne \psi (g).$