Also Indian TST 3 P1

Fun If $S$ is finite, it is clear that (2) holds, as there are infinitely many rational numbers less than 1. Therefore, let us assume that $S$ is infinite. We will prove this problem indirectly. Assume that (1), (2) both fail. Now we get that there exists a unique finite subset $S(r) \subset S$ such that $\sum_{x \in S(r)} \frac{1}{x} = r$ for each $r<1$. We will show that this is impossible, which shows the desired contradiction. We can also now easily assume that $1 \notin S$, since if there is a set $S$ which satisfies the condition and also has $1 \in S$, then $S - \{1\}$ satisfies the condition as well. With this big condition, we can force a lot of good conditions on $S(r)$. For example, if we have an element $x \in S - S(r)$, we can immediately force $S \left( r + \frac{1}{x} \right) = S(r) \cup \{x\}$, if $r+ \frac{1}{x}<1$. Let us label the elements of $S$ as $2 \le u_1 < u_2 < \cdots $. If there exists an $n$ such that $u_{n+1} < 2u_n$, let us denote $r = \frac{1}{u_n} - \frac{1}{u_{n+1}} < \frac{1}{u_{n+1}}$ Because $r<1$, there should be a unique set $S(r)$ which satisfies $\sum_{x \in S(r)} \frac{1}{x} = r$. Clearly, for any integer $x$ in $S(r)$, we will have $x > u_{n+1}$. Therefore, $u_{n+1}$ is definitely not in $S(r)$. Now, it is clear that $T = S(r) \cup \{u_{n+1}\}$ satisfies $\sum_{x \in T} \frac{1}{x} = \frac{1}{u_n}$. However, $T' = \{u_n \}$ also satisfies $\sum_{x \in T'} \frac{1}{x} = \frac{1}{u_n}$, so this shows that (1) is true. Contradiction. Now we have $u_{n+1} \ge 2u_n$ for all $n$. In this case, we can see that $\sum_{i=1}^\infty \frac{1}{u_i}$ can't be too big. Indeed, the largest possible value of $\sum_{i=1}^\infty \frac{1}{u_i}$ is $\sum_{i=1}^\infty \frac{1}{2^i} = 1$. Therefore, if $u_i \neq 2^i$ for some $i$, we immediately have $\sum_{i=1}^\infty \frac{1}{u_i} < 1$. Taking $Q = \sum_{i=1}^\infty \frac{1}{u_i} < 1$, we can just take $r = \frac{1+Q}{2}$ and see that (2) is true. Contradiction. Now we have $u_n = 2^n$ for all $n$. This is quite easy to do. For all finite subsets $T$ of $S$, the denominator of $\sum_{x \in T} \frac{1}{x}$ will be a power of 2. So we can just take something like $r = \frac{2018}{2019}$ and call it a day. Contradiction.

Assume that both assertions are false. Clearly $S$ is infinite. We first extract the uniqueness condition. Claim: There does not exists $a,b\in S$ such that $a<b<2a$. Proof: Take subset $F$ such that $\sum\limits_{x\in F}\frac{1}{x} = \frac{1}{b}-\frac{1}{a}$. Clearly $a\notin F$ by size reasons. So notice that $$\sum_{x\in F\cup\{a\}}\frac{1}{x} = \frac{1}{a}+\sum\limits_{x\in F}\frac{1}{x}=\frac{1}{b},$$contradiction to the first assertion. Enumerate $S=\{a_1,a_2,a_3,...\}$ where $a_1<a_2<...$. By the first claim $a_{i+1} > 2a_i$. Now we extract the existence condition. Claim: $a_{i+1} = 2a_i$ for any $i\in\mathbb{Z}^+$. Proof: Assume that $a_{i+1} > 2a_i$. Take any rational $r\in\left(\frac{2}{a_{i+1}},\frac{1}{a_i}\right)$ and set $F\subset S$ such that $\sum\limits_{x\in F}\frac{1}{x} = r$. Clearly $\{a_1,a_2,...,a_i\}\notin F$ for size reason. Thus $$r < \frac{1}{a_{i+1}} + \frac{1}{a_{i+2}} + \frac{1}{a_{i+3}} + ... \leqslant \frac{1}{a_{i+1}}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...\right)=\frac{2}{a_{i+1}},$$contradiction. Thus set $S$ must be in form $\{k, 2k, 4k, 8k,...\}$. However, this makes $\frac{1}{3k}$ inexpressible, final contradiction.

Loved this problem.

Functional wrote: Given any set $S$ of postive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x\in F}1/x=\sum_{x\in G}1/x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x\in F}1/x\neq r$ for all finite subsets $F$ of $S$. Solution. Assume that both the assertions do not hold. As second condition does not hold, $S$ has the infinite. Therefore for any rational $r<1$, there is an unique subset of $S$, $F$ such that $\sum_{x\in F} \frac{1}{x} = r$. Suppose there are elements $a,b$ in $S$ such that $a<b<2a$. Then there is a subset $F$ such that $\sum_{x\in F} \frac{1}{x} = \frac{1}{a} - \frac{1}{b}$. If $b\in F$, then $\frac{1}{a} = \sum_{x\in F} \frac{1}{x} + \frac{1}{b} \geq \frac{2}{b}$ which is a contradiction as we have two subsets with same sum. Suppose $S = \{a_1,a_2,\ldots \}$, and $a_i$'s are in increasing order. We have $a_{n+1} > 2a_n$ for all $n$. Therefore $a_n > 2^{n-1}a_1$. Therefore \[\sum_{i=1} \frac{1}{a_i} < \frac{1}{a_1} \left(1+\frac{1}{2} + \frac{1}{4} + \ldots \right)<\frac{2}{a_1}.\]If $a_1 \geq 2$, $\sum \frac{1}{a_1}$ is strictly less than $1$, and we can pick a rational between $1$ and that number which gives a contradiction as second condition was false. Therefore $a_1 = 1$. Now for a rational $r<1$, we have a subset $F$ such that $\sum _{x\in F}\frac{1}{x} = r<1$. Therefore $a_1\not\in F$ and again we get $\sum_{i=2} \frac{1}{a_i} < 1$ which again gives a contradiction to the second condition and we are done. $~\square$

plagueis wrote: Loved this problem. It is not a good IMO problem as it does not clearly fit into the ACGN scheme. It is not A and also not C.

prague123 wrote: plagueis wrote: Loved this problem. It is not a good IMO problem as it does not clearly fit into the ACGN scheme. It is not A and also not C. There are examples of such problems on the IMO in several places, just look at IMO 2014/5. A problem being between categories does not disqualify it, and this is a very nice problem. Besides, it is much, much more A than C, merely nonstandard A in the style of IMOSL 2016 A2, A3 or 2017 A3.

Assume for sake of contradiction that neither of the statements is true. Then, for every rational number $r\in(0,1)$, we have a unique finite set $F(r)\subset S$ such that \[r=\sum_{x\in F(r)}1/x.\]For the purposes of this assertion, having $1$ in $S$ won't affect anything since $1\not\in F(r)$ for all $r\in(0,1)$ as $r<1$, so WLOG assume $1\not\in S$. Suppose $x,y\in S$ with $x>y$. Then \[1/x+(1/y-1/x)=1/y,\]so if $x\not\in F(1/y-1/x)$ then we have two different ways of representing $1/y$ namely $\{y\}$ and $F(1/y-1/x)\cup\{x\}$. Thus, $x\in F(1/y-1/x)$, so \[1/x\le 1/y-1/x\]which means that $x\ge 2y$. So if we order $S$ as \[S=\{x_1<x_2<\cdots\},\]we must have $x_{k+1}\ge 2x_k$. This means that $x_{k+1}\ge 2^k x_1$. However, note that if $r$ can be made, then certainly \[r<\sum_{x\in S}1/x.\]This is true for $r$ arbitrarily close to $1$, so \[\sum_{x\in S}1/x\ge 1.\]But since $x_{k+1}\ge 2^k x_1$, $\sum_{x\in S}1/x\le 2/x_1$, so $2/x_1\ge 1$, so $x_1=2$. Thus, \[S=\{2,4,8,16,32,\ldots\}.\]Clearly $1/3$ is not a finite sum of these, so we have our desired contradiction.

I think it is trivial,because card(2^S)>card(Q),isn’t it correct?

@above Finite subsets of S. Cardinality of those is same as Q

Compare with https://artofproblemsolving.com/community/c6h37239p233042

@above huh this is basically the exact same problem....

Now you know why this didn't appear on a mop test

Suppose the contrary, namely that all rationals $0<r<1$ can be expressed uniquely. Clearly, this means that $S$ is infinite. Also, suppose $1\not\in S$, since this relaxes condition 1, and doesn't impact 2. Finally, define $S_r$ to be the unique subset of $S$ which generates $r$. Given $n\in \mathbb{Z}^+$, suppose that there exists $x\in S\cap[n,2n)$ such that $x\not\in S_{1/n}$. Then, the set $S_{1/n-1/x}\cup\left\{x\right\}$ also generates $1/n$, which contradicts uniqueness. Therefore, such an $x$ doesn't exist. So, all $S\cap[n,2n)\subseteq S_{1/n}$. However, the sum of reciprocals of 2 numbers in this range already exceeds $1/n$, so $|S\cap[n,2n)|\le 1$. Now, consider the intervals $[2,4)$, $[4,8)$, etc. As there is at most 1 number from $S$ in each set, $\sum\limits_{x\in S}1/x\le 1$, with equality iff $S=\left\{2^x\arrowvert x\ge 1\right\}$. Unfortunately, this choice of $S$ doesn't work, so the inequality is strict, and we can find a sufficiently small epsilon such that $1-\epsilon$ is unexpressible. This is a contradiction, as desired.

Assume that both assertion doesn't hold, that is, there exists a set $S$ of positive integers such that For every positive rational number $r < 1$, there exists a unique finite subset $F$ of $S$ such that \[ \sum_{x \in F} \frac{1}{x} = r \] Obviously $S$ is infinite, otherwise take $r < \frac{1}{\max S}$ and this contradicts the second statement. Claim 01. There doesn't exists any two element $a,b \in S$ such that $a < b < 2a$. Proof. Suppose otherwise, then we have $\frac{1}{b} < \frac{1}{a} < \frac{2}{b}$. Therefore, we can construct $\frac{1}{a}$ in two ways: which is just the partition of $\frac{1}{a}$, which exists by the hypothesis, or $\frac{1}{b}$ and the partition of $\left( \frac{1}{a} - \frac{1}{b} \right)$, a contradiction. Now, consider the elements of $S$ ordered from the smallest number: $a_1 < a_2 < a_3 < \dots$ By Claim 01, we have $a_{i + 1} \ge 2a_i$ for every $i \in \mathbb{N}$. We could WLOG $a_1 \not= 1$, as whether $1$ is in or not won't affect the problem as the hypothesis only holds for $r < 1$. Claim 02. $a_1 = 2$. Proof. Suppose otherwise, that $a_1 = 3$. Then \[ \sum_{x \in S} \frac{1}{x} \le \sum_{i = 0}^{\infty} \frac{1}{2^i \cdot 3} = \frac{2}{3} \]Taking $r > \frac{2}{3}$, this clearly won't satisfy. In a similar method, one can prove that $a_i = 2^i$ for every $i \in \mathbb{N}$. Now, we'll prove that $r = \frac{1}{3}$ is not attainable. To prove this, we claim that the denominator of $r$ will stay $0$ modulo $3$ at the whole process of subtraction. \[ r - \frac{1}{2^k} = \frac{1}{3 \cdot N} - \frac{1}{2^k} = \frac{2^k - 3 \cdot N}{3 \cdot N \cdot 2^k} \]It's clear enough that $2^k - 3 \cdot N$ is not $0$ modulo $3$, and therefore we've finished.

In my proof, I implicitly use the fact that $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges. Assume $(2)$ doesn't hold. Then, clearly $S$ is an infinite set of positive integers. Given two finite subsets $A$ and $B$ of $S$, and WLOG $\sum_{x \in A}^{} \frac{1}{x} > \sum_{x \in B}^{} \frac{1}{x}$. It suffices to prove $$\sum_{x \in A}^{}\frac{1}{x}=r \text{ and } \sum_{x \in B}^{}\frac{1}{x}=r-\epsilon \text{ such that }\epsilon<\text{ min }{\frac{1}{x} : x \in A \cup B}$$Consider successively deleting $\frac{1}{l}$ from $\sum_{x \in A}^{}\frac{1}{x}$ where $l$ is the maximal element of $A$, and let $A$ now be the new sequence of integers obtained from deleting $l$. We do this until $\sum_{x \in A \setminus{j}}^{} \frac{1}{x} < \sum_{x \in B}^{} \frac{1}{x}$ where $j$ is the maximum element set $A$. We consider our current sets $A$ and $B$ from now on. Like in the above notation, let $\sum_{x \in A}^{} \frac{1}{x}-\sum_{x \in B}^{} \frac{1}{x}=\epsilon$. Furthermore, let $j$ be the maximal element of $A$. Then, $\sum_{x \in A \setminus{j}}^{} \frac{1}{x} < \sum_{x \in B}^{} \frac{1}{x} \implies \epsilon=\sum_{x \in A}^{} \frac{1}{x}-\sum_{x \in B}^{} \frac{1}{x}< \frac{1}{j}$. Then, let $w$ be the maximal element of $B$ and consider the set $B'=\{u \in \mathbb{Z}: u>w\}$. Now, consider adding the least remaining element of $B'$, $m$ for convenience to $B$, and call the new set $B \cup \{m\}$ as $B$, and let $\sum_{x\in A}^{}\frac{1}{x}-\sum_{x\in B}^{}\frac{1}{x}=\epsilon$ as before. We keep applying the above operation to $G$ until $\epsilon>0$ and applying the operation $1$ more time results in $\epsilon<0$. Let the number we add to $G$ such that $\epsilon<0$ be $l$. Then $$\sum_{x \in A}^{} \frac{1}{x}-\sum_{x\in B}^{} \frac{1}{x}-\frac{1}{l}<0 \implies \epsilon-\frac{1}{l}<0 \implies \epsilon<\frac{1}{l}$$However, $\epsilon<\{\frac{1}{x}:x \in A \} \text{ and } \epsilon<\frac{1}{l}<\{\frac{1}{x}: x \in B \}$. Therefore, given $(2)$ does not hold, we can find finite subsets $A$ and $B$ of $S$ such that $\sum_{x \in A}^{}\frac{1}{x}=r \text{ and } \sum_{x \in B}^{}\frac{1}{x}=r-\epsilon \text{ such that }\epsilon<\{\text{ min }{\frac{1}{x} : x \in A \cup B}\}$, and there must exist a finite subset $L$ of $S$ completely distinct from $F$ and $G$ such that $$\sum_{x\in L}^{} \frac{1}{x}=\epsilon \implies \sum_{x \in A}^{} \frac{1}{x}=\sum_{x \in B \cup L}^{} \frac{1}{x}$$ Otherwise, $(2)$ holds, and we are done.

Solved with eisirrational and Th3Numb3rThr33. Assume for contradiction both fail. Then for each $r<1$ there is a unique finite subset $R(r)$ of $S$ such that $\textstyle\sum_{x\in F}1/x=r$. Claim: There are no two elements $a,b\in S$ with $a<b<2a$. Proof. Let $a,b\in S$ with $a<b$, so that $R(1/a)=1/a$ and $R(1/b)=1/b$. Note that if \[\frac1b\notin R\left(\frac1a-\frac1b\right),\]then $R(1/a-1/b)\cup\{1/b\}$ is another way to represent $1/a$, contradiction. Hence \[\frac1a-\frac1b\ge\frac1b\implies b\ge2a,\]as claimed. $\blacksquare$ Now if $S$ is finite, then (ii) clearly holds, and if $1\in S$, we may delete it since it is not $R(r)$ for all $r<1$. Thus we have \[\sum_{x\in S}\frac1x\le\frac12+\frac14+\frac18+\cdots=1,\]where equality holds if and only if $S=\{2,4,8,\ldots\}$. If equality fails, then some $r<1$ cannot be represented, but if $S=\{2,4,8,\ldots\}$, then it is impossible to represent $1/3$, the end.

There are many similar question.

Indeed: Problem $6$ from this has a similar taste.

Functional wrote: Given any set $S$ of postive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x\in F}1/x=\sum_{x\in G}1/x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x\in F}1/x\neq r$ for all finite subsets $F$ of $S$. Assume to the contrary such a set exists. We begin by observing that $S$ must be infinite. Now suppose that $S=\{a_1,a_2,\ldots\}$, where $a_{i+1}>a_i$. Also observe that $1\not\in S$. If we had $\frac{1}{a_i}-\frac{1}{a_{i+1}}< \frac{1}{a_{i+1}}$, then we are violating (2). Then we must have $\frac{1}{a_i}\ge \frac{2}{a_{i+1}}\implies a_{i+1}\ge 2a_i$. Thus we must have $a_i\ge 2^i$. Suppose $N$ is the smallest positive integer such that $a_N>2^N$. Then we have $$\sum_{x\in S}<1$$This again violates (2). But if $S=\{2,4,8,\ldots\}$, we can not represent $\frac{1}{3}$ using finitely many elements of $S$, a contradiction.

Really nice problem! If $S$ is a finite set, there are only a finite number of subsets of $S$; hence, as there are an infinite number of rational numbers $r < 1$, condition (2) is true. Henceforth, we assume $S$ is an infinite set. Suppose condition (2) is false, i.e. for each rational number $r < 1$, there exists a finite subset $F_r$ of $S$ such that $\sum_{x \in F_r} \frac{1}{x} = r$. In what follows, we assume that $1$ is not an element of $S$; since $1 \not\in F_r$, removing $1$ from $S$ if it exists will not alter the truth of condition (2). We will show that condition (1) is true. Main Claim: There exist $u, v \in S$ for which $u > v > \frac{u}{2}$. Proof: Suppose not, and let the elements of $S$ be $1 < a_1 < a_2 < \cdots$. For each $i > 1$, we have $\frac{a_{i + 1}}{a_i} \geq 2$, so $a_k \geq 2^{k - 1}a_1$. Hence, we have \begin{align*} T &= \sum_{i = 1}^\infty \frac{1}{a_i} \\ &\leq \sum_{i = 1}^{\infty} \frac{1}{2^{i - 1}a_1} \\ &\leq \frac{1}{a_1}\sum_{i = 0}^\infty \frac{1}{2^i} \\ &\leq \frac{2}{a_1} \\ &\leq 1. \end{align*}Equality holds iff $a_k = 2^k$ for each $k$. If $T < 1$, pick a rational number $r$ for which $1 > r > T$. We have \begin{align*} r = \sum_{x \in F_r} \frac{1}{x} < \sum_{x \in S} \frac{1}{x} = T < r, \end{align*}contradiction. Hence, we must have $T = 1$, implying that $a_k = 2^k$ for each $k$. However, as $S$ is the set of powers of $2$ larger than $1$, there is no finite subset $F \in S$ such that $\sum_{x \in F} \frac{1}{x} = 0.\overline{1}_2 = \frac{1}{3}$, contradicting the existence of $F_{\frac{1}{3}}$. Thus, we have reached a contradiction, proving the claim. $\blacksquare$ Now, consider $u, v \in S$ for which $u > v > \frac{u}{2}$. Note that $\frac{1}{v} - \frac{1}{u} < \frac{1}{u}$, so $F_{\frac{1}{v} - \frac{1}{u}}$ does not contain $u$. Let $F = F_{\frac{1}{v} - \frac{1}{u}} \cup \{u\}$. We have $F \neq \{v\}$, and $\sum_{x \in F} \frac{1}{x} = \sum_{x \in \{v\}} \frac{1}{x}$. Thus, condition (1) is true, so we are done. $\Box$

FTSoC assume both are false. Let $s(\mathcal{F})$ for every $\mathcal{F} \subseteq S$ be the sum of $\tfrac 1x$ over all $x \in \mathcal{F}$. Then we have $s$ is injective, and surjective on $(0, 1)$. We claim, that we can actually uniquely define $S$. Let $x, y$ be elements of $S$. WLOG $1 \not \in S$ since we are only given the surjective condition on $< 1$, so this $1$ term would not matter. WLOG $x > y$, then if $\tfrac 1y - \tfrac 1x \in (0, 1)$ note that $x$ must be in the set $F$ for which $s(F) = \tfrac 1y - \tfrac 1x$, else there are two ways of representing $s(F)$, by simply using $y$ or using both $F$ combined with $x$. So this tells is $x \geq 2y$ by bounding. Next, we rank $S$'s elements in order $a_1 < a_2 < \ldots$ where by the previously derived condition, clearly $a_{i+1} \geq 2a_i$. We will show that in fact, equality must hold. If not, then choosing $r = 1 - \epsilon$ cannot be achieved. Hence indeed, we can directly classify $S$ as the set of all powers of $2$ at least $2$. But this set clearly is not surjective, contradiction. $\blacksquare$

Also Thailand TST 2019

assume that both statement fail For each r there exists F such that the sum of the reciprocal of the numbers in F is r . Suppose F={a1,a2,……aj} obviously, we have S is infinite, othwise statement 2 holds. (1) if there exists an number m such that r/2<m≤r and m is not in F use statement 2, we have a subset F1 such that the sum of the reciprocal of the numbers in F is r-m notice that r-m < r/2 < m so m is not in F1 regard F1∪{m} and F, we have a contradiction. (2) else, suppose a is in S, regard the region (a,2a) let r be 2a by using (1), we have that there's no number in that region which is also in S using the same method, we have: The maximum of the sum of the reciprocal of the numbers in S and ≤a is a+2/a+4/a+…… let r be 2a-ε so 2/a,4/a,…… is all the number in S and <a but since all numbers in S are intigers, we have a contradiction.

Let $A$ be the set $\{1/x \mid x \in S\}$. Clearly, $A$ contains an infinite number of elements. Let the elements of $A$ be $a_1 \ge a_2 \ge a_3 \ge \dots$ For some $r \in \mathbb{Q}$, let $A(r)$ denote the set of elements in $A$ which sum to $r$. FTSOC, assume that both conditions are false. Then, from condition (1) we have that all elements of $A$ are distinct. Furthermore, from (2) we have the following: Claim: $a_{i+1} \le a_{i}/2$ Proof. Otherwise, we can choose $r$ such that $2a_i > 2a_{i+1} > r > a_i$. Then, $r-a_i < a_i$ and $r-a_{i+1} < a_{i+1}$ so the distinct sets $a_i \cup A(r-a_i)$ and $a_{i+1} \cup A(r-a_{i+1})$ both sum to $r$, contradiction. $\square$ As we are considering $r<1$, we can assume that $a_1 \le 1/2$. Then the previous claim implies that $a_2 \le 1/4$, $a_3 \le 1/8$, and so on. If one of the preceding inequalities is strict, then the sum of all elements of $A$ is less than 1, contradicting (2). Otherwise, the elements of $A$ must be $1/2, 1/4, 1/8, 1/16, \dots$, so numbers like $1/7$ cannot be written as a sum of distinct elements of $A$, contradiction.

Suppose both statements' negations are true. In other words, every rational number $r<1$ can be expressed as $\sum_{x\in F} 1/x = r$ for some finite subset $F$ and such subset is unique. Claim: There cannot exist $a,b\in S$ with $a<b < 2a$. Proof. By assumption there then exists subset $F$ with $$\sum_{x\in F} 1/x = \frac1a - \frac1b < \frac 2b - \frac 1b = \frac 1b < \frac{1}{a},$$so $a,b\not\in F,$ implying that $$\sum_{x\in F\cup\{b\}} = \frac{1}{a},$$Contradiction. $\blacksquare$ Claim: For all $x>1$ with $x\in S,$ $\sum_{x\in S} 1/x \ge 1.$ Proof. Suppose not, and that $\sum_{1<x \in S} 1/x = r < 1,$ then $r+\varepsilon < 1$ for sufficiently small $\varepsilon \in \mathbb Q$ cannot be obtained, contradiction. $\blacksquare$ However, it is obvious then from our first claim that, $$\sum_{1< x \in S} 1/x \le \sum_{i=1}^\infty \frac{1}{2^i} = 1,$$So from our earlier claim, this forces $S = \{2,4,8,\cdots\}.$ Consequently, each rational $r<1$ is uniquely expressible as its binary representation. However, $2/3 = 0.\overline{10}_2$ is infinitely periodic, meaning no such finite subset of $S$ has elements' sum of $2/3,$ contradiction. Therefore, we conclude at least one of the statements must be true. $\blacksquare$

Functional wrote: Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x\in F}1/x=\sum_{x\in G}1/x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x\in F}1/x\neq r$ for all finite subsets $F$ of $S$. FTSoC such a set $S$ exists which satisfied both the conditions. The second condition implies that $|S|$ is infinite. If there exists $x, y \in S$ such that $y \in [x, 2x]$, then we see that choosing elements $s_1, s_2 \dots s_k \in S$ such that $\sum\limits_{i=1}^k s_i = \frac{1}{x} - \frac{1}{y} < \frac{1}{y}$, then $\sum\limits_{i=1}^k s_i + \frac{1}{y} = \frac{1}{x}$, a contradiction to condition $2$. We now remove $1$ from $S$ as it is not useful to $S$ now. We see that $\sum\limits_{x \in S} \frac{1}{x} \leq \sum\limits_{i=1}^{\infty} \frac{1}{2^i} = 1$, which means that if $S \neq \{ 2, 4, 8 \dots \}$ then there exists arbitrarily small positive reals $\epsilon$ for which $1 - \epsilon$ cannot be represented as $\sum\limits_{i=1,x_i\in S}^k \frac{1}{x_i} = 1-\epsilon$ and if $S = \{2, 4, 8 \dots \}$ then any fraction $\frac{p}{q}$ with $\text{rad}(q) \neq 2^k$ for some $k \in \mathbb{Z}^+$ cannot be represented as in the second condition, a contradiction. Therefore no such set $S$ which satisfies both the conditions.

Assume that both conditions are not satisfied. Let $s$ be the smallest element of $S$. Suppose that there is an $r \in S$ such that $s<r<2s$. Now, observe that there exists a subset $F$ of $S$ such that $\frac{1}{s}-\frac{1}{r}= \sum_{x \in F} \frac{1}{x}$. Observe that if $r \not \in F \implies \frac{1}{s}= \sum_{x \in F \cup \{r\}} \frac{1}{x}$, a contradiction since $s \in S$. Thus, $r \in F \implies \frac{1}{2s} > \frac{1}{s}-\frac{1}{r} \geq \frac{1}{r}> \frac{1}{2s}$, contradiction. Now, let $r \geq 2s$ be the smallest element of $S$ greater than $s$. Then, $$\frac{1}{2s} \leq \frac{1}{s}-\frac{1}{r}= \sum_{x \in F} \frac{1}{x}$$for some subset $F$ of $S$. If $r \not \in F$, then $\frac{1}{s}=\sum_{x \in F \cup \{r\}} \frac{1}{x}$, a contradiction, since $s \in S$. Thus, $r \in F \implies \frac{1}{r} \geq \frac{1}{s}-\frac{1}{r}= \sum_{x \in F} \frac{1}{x} \geq \frac{1}{r} \implies \frac{1}{s}-\frac{1}{r}=\frac{1}{r} \implies r=2s$. By the same argument, we can prove that $S=\{s,2s,2^2s,2^3s,...\}$. If $s>2$, then $\sum_{x \in S} \frac{1}{x}$ converges to a number less than $1$, so we can take a rational number in the interval $(\sum_{x \in S} \frac{1}{x},1)$ ($\mathbb{Q}$ is dense in $\mathbb{R}$) to reach a contradiction. Thus, $s=1$ or $s=2$. If $s=1$, we can simply ignore $s$ from $S$, so assume WLOG that $s=1$. But then $\frac{1}{3}$ cannot be represented as $\sum_{x \in F} \frac{1}{x}$ where $F$ is a finite subset of $S$, so we are done. $\blacksquare$

Let both of the statements be false. We have that for each $r\in (0,1)$ there exists a unique subset $S(r)$ such that \[\sum_{x\in S(r)}\left(\frac{1}{x}\right)=r\]Clearly $S$ must not be finite, since then there is a finite number of possible $S(r)$ but an infinite number of rational numbers, and there cannot be a bijection there. Now, suppose there exist $a,b\in S$ such that $a<b<2a$ then \[\frac{1}{a}-\frac{1}{b}\le \frac{1}{b}\]which means that $b\notin S(\tfrac{1}{a}-\tfrac{1}{b})$ so $\{a\}=S(\tfrac{1}{a})=S(\tfrac{1}{a}-\tfrac{1}{b})\cup \{b\}$, impossible. Now, let any two consecutive terms of $S$ arranged in increasing order be $a>b$. If $a>2b$ then it is easy to see that there is no way to represent $r\in(\tfrac{2}{a},\tfrac{1}{b})$. If we have nothing smaller than $a$ then the maximum we can get is \[\frac{1}{a}+\frac{1}{2a}+\frac{1}{4a}+\dots = \frac{2}{a}\]and if we have something smaller than $a$ then the minimum we can get is $\tfrac{1}{b}$. Therefore, we must have \[S=\{a,2a,4a,8a,\dots\}\]and the maximum sum will be $\tfrac{1}{a}$ so $a=1$. Since we have found $S$, we can very confidently say that there are so many rational numbers that don't have a corresponding subset.

Amazing! Assume that neither assertion holds. Then suppose FTSOC that $x,y\in S$ satisfy $x<y<2x$. By considering \[\frac{1}{x}=\frac{1}{y}+\left(\frac{1}{x}-\frac{1}{y}\right)\]we can actually generate two different subsets of $S$, violating the first condition. However we also know that if we sum the reciprocals of all elements $x\in S$ satisfying $x>1$ we obtain a value which is at least $1$ (else we violate the second condition). This implies that these elements are exactly the powers of $2$. However by considering base $2$ we find that $\frac{1}{3}$ isn't achieveable so we are done. $\blacksquare$

Let, $S = \{x_1, x_2, x_3, \dots\}$ where, $x_1 < x_2 < \dots < x_n < x_{n+1} < \dots$ Now, there are two cases, $\textbf{Case 1:}$ There exists $k$ s.t. $x_{k+1} < 2x_k$ If the second condition doesn't hold, for all $1 > r \in \mathbb Q$ we will get a finite subset $F_r$ that gives the reciprocal sum $r$. Now, take $r = \dfrac{x_{k+1} - x_k}{x_kx_{k+1}}$. Then by our condition we get $r < \dfrac1{x_{k+1}}$. So, $F_r$ doesn't contain any $x_n$ with $n \leq k+1$. Now take, $F = \{x_k\}$ and $G = \{x_{k+1}\} \cup F_r$ and we get our first condition true. $\blacksquare$ $\textbf{Case 2:}$ For all $k$ we have $x_{k+1} \geq 2x_k$ We want to show that the second condition holds. Notice that, For the second condition it doesn't matter if $1$ is in $S$ or not. So we assume that $1 \notin S$. Then, $x_1 \geq 2$ and if we get $x_{k+1} > 2x_k$ at some point then, let $c = \dfrac1{2x_k} - \dfrac1{x_{k+1}}$ and we will get that the total sum of reciprocals of $S$ is at most $1-c$. Then we can just take a $r = 1-c + \epsilon$ for small enough $\epsilon > 0$ and we will be done. And similarly if $x_1 \neq 2$ we get a bound $\frac2{x_1}$ on the total sum. So now the remaining case is $x_i = 2^i$ for all $i$. But then we take $r = (0.\overline{01})_2 = (1/3)_{10}$ and if we get a set $F_r$ for this then we will get a finite representation of $r$ in binary which we can easily show that doesn't exist. So we are done. $\blacksquare$

Assume not. Note that $S$ is infinite because otherwise there is only finitely many sums attainable but there is infinite rationals between $0$ and $1$. Enumerate the elements of $S$ as $s_1 < s_2 < \dots$. Let $p(F) = \sum_{x \in F} \frac{1}{x}$. Then for all rational $r<1$ there is a finite set $F \in S$ such that $p(F) = r$ and for all pairs of finite sets $F,G \in S$, $p(F) \neq p(G)$. Let's assume that $s_{k+1} < 2s_k$ for some $k$. Then consider the assertion of ($2$) for $\frac{1}{s_k}-\frac{1}{s_{k+1}}$. Note that this value is less than $\frac{1}{s_{k+1}}$, so the finite set $F$ satisfying the condition only contains terms greater than $s_{k+1}$. Let $F$ be the satisfying set. Then $p\left(F \cup \{s_{k+1}\} \right) = \frac{1}{s_k} = p\left(\{ s_k \}\right)$, a contradiction to $(1)$ and so we're done with this case. Thus $s_{k+1} \ge 2s_k$ for all $k$. Consider a construction for ($2$). Note that $1$ isn't in it as that would make it too big. WLOG $s_1>1$ (if it's $1$ just chop it off). Note that $\sum_{x \in S} \frac{1}{x}$ converges since it's bounded by $\sum_{i=1}^{\infty} \frac{1}{2^i} = 1$. If $s_{k+1} > 2s_k$ for some $k$, $\sum_{x \in S} \frac{1}{x} = \sum_{i=1}^{k} \frac{1}{s_i} + \sum_{i=k+1}^{\infty} \frac{1}{s_i} \le \sum_{i=1}^k \frac{1}{2^i} + \sum_{i=k+1}^{\infty} \frac{1}{2^i+1} < \sum_{i=1}^{\infty} \frac{1}{2^i} = 1$. Thus if we let $r = \sum_{x \in S} \frac{1}{x}$, we won't be able to find a set summing to that (they would all be smaller). Thus $s_{k+1}=2s_k$ and $s_k=2^k$. We finish by taking the assertion for $r=\frac{1}{3}$ and noting that the denominator in any sum would be a power of $2$, which $3$ obviously isn't. This provides a contradiction and concludes the proof.

Let $s(T)$ denote the sum of the reciprocals of some set $T$. Suppose that neither of these conditions hold. Clearly $S$ is infinite, and let its elements other than $1$ be $1<a_1<a_2<\cdots$. The key claim is the following. Claim: We need $a_{i+1} \geq 2a_i$ for all $i$. Proof: Suppose otherwise. Then there exists some finite $T \subset S$ such that $s(T)=1/a_i-1/a_{i+1}<1/a_{i+1}$, so $a_{i+1} \not \in T$. But then $s(T \cup \{a_{i+1}\})=s(\{a_i\})$: contradiction. $\blacksquare$ This implies that $\sum_{x \in S} 1/s$ converges and is at most $\sum_{i \geq 1} 1/2^i=1$, with equality iff $S=\{2,4,8,\ldots,\}$. But it must be exactly $1$ as well, since we should be able to form numbers arbitrarily close to $1$, so we need equality. But then we can only form dyadic rationals, so this contradicts condition 2. $\blacksquare$

Suppose not. Redefine $S$ as a set of reciprocals of whole numbers. Then we can define a function $f(x)$ from rational numbers less than $1$ to finite subsets of $S$ such that $f(x)$ is the unique finite subset whose elements add to $x$. For the rest of the argument I will give, assume WLOG that $1$ is not in $S$ (this does not change the conclusion). Let $s_i$ denote the $i$th largest element in $S$. Observe that $f(s_i)=\{s_i\}$. This means that for $s_i > s_j$, we have $s_j \in f(s_i-s_j)$. Otherwise we could write $s_i=\sum f_(s_i-s_j) \cup \{s_j\}$. A similar argument shows $s_j \notin f(s_i-2s_j)$. We can repeat this. In general, $s_j \in f(s_i-(2n+1)s_j)$ for an integer $n$. Thus, for integers $n$, if $s_i-(2n+1)s_j>0$, then $s_i-(2n+1)s_j \ge s_j$. This means that $\lfloor \frac{s_i}{s_j} \rfloor$ is even for all $i, j$. This implies that if $a \in S$, the next smallest element that could possible be in $S$ is $\frac{a}{2}$. Thus the set $S$ which achieves maximal total sum is the set consisting of fractions which are negative powers of $2$, which has sum exactly $1$. If the set $S$ is not this set, then its elements sum to less than $1$, so we can find a contradiction by finding a rational number $p$ such that $p$ is bigger than the sum of $S$, whence $f(p)$ is undefined. If our set $S$ is the set of negative powers of $2$, then $f(\frac{1}{3})$ is not defined, because of the requirement that the subsets be finite, and $\frac{1}{3}$ in binary is $0.0111\ldots$.

Assume FTSOC there exists an $S$ that doesn't satisfy either assertion. Clearly $S$ must have infinite cardinality. Label the elements of $S$ as $a_1 < a_2 < a_3 \dots$. Notice that for any $x,y \in S$ and $x>y$, there must exist a subset $F$ that has the reciprocals of its elements sum to $\frac{1}{y} - \frac{1}{x}$. Now if $x \notin F$, then when we simply add $x$ to $F$, this new set will have its sum be equal to $\frac{1}{y}$ which is it not possible due to the 1st assumption as we can simply take the set $\{ y \}$. Therefore $x$ must be in $F$, which implies that $\frac{2}{x} \le \frac{1}{y}$, and $x \ge 2y$. Therefore if we have two elements $x > y$, then we must have $x \ge 2y$. From this we easily get $a_{i+1} \ge 2 \cdot a_i$. Now notice that $$\sum_{x \in S} \frac{1}{x} \le \sum_{i=0}^{\infty} \frac{1}{2^i \cdot a_{1}} = \frac{2}{a_{1}}$$Now we have $$\sum_{x \in S} \frac{1}{x} \ge 1$$because if not, for all $F \subset S$ we will have $\sum_{x \in F} \frac{1}{x} < 1$, which contradicts second assumption. Now this simply implies that $a_{1} = 1, 2$ and that $a_{i+1} = 2 \cdot a_{i}$. Now simply take a rational with denominator not a power of $2$. This set cannot achieve such a rational, so we have achieved a contradiction.

Nice problem. It looks like a difficult problem, but it isn't. I have solved this in the same way as (4) Therefore not uploading

AFTSOC that both statements are true. Claim: $\frac{1}{s_1} = \sum_{i\ge2}\frac{1}{s_i}$. If $\frac{1}{s_1} > \sum_{i\ge2}\frac{1}{s_i}$, then there exists a rational number $r$ between $1$ and $\sum_{i\ge1}\frac{1}{s_i}$ since $\sum_{i\ge1}\frac{1}{s_i} < \frac{2}{s_1} \le 1$ and the rationals are dense in the reals. If $\frac{1}{s_1} < \sum_{i\ge2}\frac{1}{s_i}$, then there must exist a $k$ such that $\frac{1}{s_1} < \sum_{i=2}^k \frac{1}{s_i}$ and $\frac{1}{s_1} > \sum_{i=2}^{k-1} \frac{1}{s_i}$. Now by assumption, there exists a set $A$ such that $\sum_{a\in A} \frac{1}{s_i} = \left(\sum_{i=2}^k \frac{1}{s_i}\right) - \frac{1}{s_1} < \frac{1}{s_k}$. So all elements of $A$ must be less than $k$, so we can just take $\{1\} \cup A$ and $\{2, \dots, k\}$ to get a contradiction, so we have proven the claim. Now note that $s_1= 2$ otherwise $r = 1 - \epsilon$ gives a contradiction. Now this argument clearly works for $s_2$ as well, by only considering numbers under $\frac{1}{2}$ so we get that $s_i = \frac{1}{2^i}$. Now taking $r = \frac{1}{3}$ gives a contradiction so we are done.

The problem is trivial if $S$ is finite, so suppose it's infinite, FTSOC assume that neither of the claims was true, then trivially all elements of $S$ are distinct from each other so we let $S= \{a_1, a_2, \cdots \}$ where we have them ordered in increasing order, meaning that $a_i<a_{i+1}$ and also note that $a_1 \ge 2$ since if a set had $1$ and satisfied the condition then so will $S-{1}$, so it doesn't really matter. First we take two elements $a_k>a_j$ (so of course $k>j$), then consider $\sum_{x \in S_{k,j}} \frac{1}{x}=\frac{1}{a_j}-\frac{1}{a_k}$, suppose FTSOC that $2a_j>a_k$, then it's clear that in $S_{k,j}$ a subset of $S$ we can't have $a_k$ on it for size reasons, so now note that we have $\sum_{x \in S_{k,j} \cup (a_k)} \frac{1}{x}=\frac{1}{a_j}$ which contradicts the opposite of (1), therefore we have $a_k \ge 2a_j$, this in general implies that $a_{i+1} \ge 2a_i$ for any $i \in \mathbb Z_{>0}$. Now suppose FTSOC that we have some $i$ for which $a_{i+1}>2a_i$, then consider a rational number $r$ strictly between $\frac{2}{a_{i+1}}$ and $\frac{1}{a_i}$, then let $S_1$ be the finite subset such that $\sum_{x \in S_1} \frac{1}{x}=r$, notice that for size reasons we can't have any $a_1, \cdots a_i$ on $S_1$ therefore $\sum_{x \in S_1} < \frac{1}{a_{i+1}} \left(1+\frac{1}{2}+\cdots \right)=\frac{2}{a_{i+1}}$ which contradicts the existence of $r$, therefore $a_{i+1}=2a_i$ must be true for all $i \in \mathbb Z_{>0}$. Therefore we have that $S$ is of the form $(\ell, 2\ell, 4\ell, \cdots )$ but now it's impossible to create $\frac{2}{\ell}$ for all $\ell \ge 3$ meaning that either $\ell=1$ or $2$, but since we can get rid of $1$ we can just consider $\ell=2$ in which case we have $S$ of the form $(2, 4, 8 \cdots)$ in which from $v_2$ and base $2$ we can check that it happens to be impossible to make $\frac{1}{3}$, thus we are done .

Suppose ftsoc that a set $S$ doesn't satisfy any of the two conditions. Clearly, $S$ must be infinite. For a rational number $r\ge 0$, call its decomposition $F\subset S$ if $\sum_{x\in F}\frac{1}{x}=r$. Any rational number $r<1$ has a unique decomposition, and no rational number has more than one decomposition. We start with a fairly simple observation : Claim : For any element $s\in S$ and any rational number $r\in \left[\frac{1}{s}; \frac{2}{s}\right[$, $r$ has a unique decomposition, which contains $s$. Moreover, if any rational number $r\in \left[\frac{2}{s}; \frac{3}{s}\right[$ has a decomposition, it does not contain $s$. Proof : For the first part : note that $r-\frac{1}{s}\in \left[0;\frac{1}{s}\right[$. Therefore $r-\frac{1}{s}$ has a unique decomposition, and it cannot contain $s$. It follows that $r$ has a decomposition which contains $s$. Now, suppose some $r\in \left[\frac{2}{s}; \frac{3}{s}\right[$ has a decomposition which contains $s$. Then, $r-\frac{1}{s}\in \left[\frac{1}{s}; \frac{2}{s}\right[$ has a decomposition which doesn't contain $s$, which is a contradiction. $\square$ This leads to the following claim : Claim : Suppose $S=\{a_1,a_2,\ldots\}$ where $a_1<a_2<\ldots$. Then, for each positive integers $k$ and $n$, we have $a_{k+1}\ge 2a_k$ and $\frac{1}{a_k}>\sum_{i=1}^{n}\frac{1}{a_{k+i}}$. Proof : Note that $r=\frac{1}{a_{k}}+\frac{1}{a_{k+1}}>\frac{2}{a_{k+1}}$ is a rational number which contains $a_{k+1}$ in its decomposition. Therefore, it must be the case that $r\ge \frac{3}{a_{k+1}}$ by our claim, which gives $a_{k+1}\ge 2a_k$. Now, suppose that some $k$ and $n$ are such that $\frac{1}{a_k}\le \sum_{i=1}^{n}\frac{1}{a_{k+i}}$, and take the minimal such $n$. We then have $\sum_{i=1}^{n}\frac{1}{a_{k+i}}\in \left[\frac{1}{a_k}; \frac{2}{a_k}\right[$, which contradicts our claim since its decomposition must contain $a_k$. $\square$ Let us fix some $k\in \mathbb N$. The sequence $x_n=\sum_{i=1}^{n}\frac{1}{a_{k+i}}$ is increasing and bounded above, so it must have a limit $l=l(k)$, which must not be larger than $a_k$ by our claim. If $l<\frac{1}{a_k}$, then any rational number $q\in \left]l;\frac{1}{a_k}\right[$ does not have a decomposition. This is a contradiction, since every number in this interval is smaller than $1$. Therefore, we must have $l(k)=\frac{1}{a_k}$. Since $a_{k+i}\ge 2^ia_k$, it must be the case that $a_{k+i}=2^ia_k$ for every $i$ and $k$. If not, then clearly $l(k)<\frac{1}{a_k}$ by an easy geometric series argument. In particular, we have $a_k=2^{k-1}a_1$ for every $k$. This cannot hold, since $\frac{1}{3a_1}$ would not have a decomposition. $\blacksquare$