Let $A$ and $B$ infinite sets of positive real numbers such that: 1. For any pair of elements $u \ge v$ in $A$, it follows that $u+v$ is an element of $B$. 2. For any pair of elements $s>t$ in $B$, it follows that $s-t$ is an element of $A$. Prove that $A=B$ or there exists a real number $r$ such that $B=\{2r, 3r, 4r, 5r, \dots\}$.
Problem
Source: 2024 Mexican Mathematical Olympiad, Problem 5
Tags: 2024, Mexico, algebra, Reals
KAME06
07.11.2024 05:00
Sea el conjunto $A=\{a_1, a_2, ...\}$, donde $a_1<a_2<a_3<....$
1)Por hipótesis, $a_1+a_1=2a_1 \in B$ y $a_1 + a_k \in B$, $k>1$ entero.
2)Por hipótesis y lo anterior, $a_1 + a_k - 2a_1=a_k-a_1 \in A$
3)Por hipótesis, $a_k - a_1 + a_1 = a_k \in B$
4)Usando 2), $a_2-a_1 \in A$, pero este valor claramente es menor que $a_k$, por tanto, $a_2-a_1=a_1\Rightarrow a_2=2a_1$
Ahora, por inducción fuerte en $n$, demostremos que $a_n=na_1$ para todo $n>1$
Caso base: n=2 hecho
Paso inductivo: si $a_i=i * a_1$ para toda $i=1, 2, ..., n-1$, notemos que, por 3), $a_n, a_1 \in B$, por lo que por hipótesis, $a_n-a_1\in A$.
Por tanto $a_n-a_1=a_t=t*a_1 \Rightarrow a_n=(t+1)a_1$ para algún $t$ menor que $n$. Si $t<n-1$, $a_n=0$ o $a_n=a_{t}$, lo cual es en ambos casos absurdo, por tanto, $t=n-1$ y $a_n=n*a_1$.
5)Ya demostrado eso, supongamos que existe un $b<a_2$ tal que $b\in B$. Por 3), $a_2-b\in A$, y por tanto $a_2-b=a_1 \Rightarrow b=a_1$ De aquí que el menor elemento de $B$ solo podría ser $a_1$.
Por tanto, $a_1\in B$. Si $u \in B$, $u>a_1$ tenemos que $u-a_1\in A$, por 4), $u-a_1=a_m\Rightarrow u=m*a_1+a_1=(m+1)a_1=a_{m+1}$, para algún $m$
Por 5) y 3) Concluimos, en este caso, que $A=B$
6) Si el mínimo valor de un elemento de $B$ fuera $a_2$, si $u \in B$, $u>a_2$ tenemos que $u-a_2\in A$, por 4), $u-a_2=a_m\Rightarrow u=m*a_1+a_2=m*a_1+2*a_1=(m+2)a_1=a_{m+2}$, para algún $m$
Por 3) y 4), de aquí sacamos que $B=\{2a_1, 3a_1, 4a_1, ...\}$.
Let the set $A=\{a_1, a_2, ...\}$, where $a_1<a_2<a_3<....$
1)We have that, $a_1+a_1=2a_1 \in B$ and $a_1 + a_k \in B$, $k>1$ integer.
2)We have that, $a_1 + a_k - 2a_1=a_k-a_1 \in A$
3)We have that, $a_k - a_1 + a_1 = a_k \in B$
4)Using 2), $a_2-a_1 \in A$, but this is less than $a_k$, so, $a_2-a_1=a_1\Rightarrow a_2=2a_1$
Using strong induction on $n$, we are going to show that $a_n=na_1$ for all $n>1$
Base case: n=2 done
Inductive step: If $a_i=i * a_1$ for all $i=1, 2, ..., n-1$, note that, using 3), $a_n, a_1 \in B$, so, $a_n-a_1\in A$.
Therefore $a_n-a_1=a_t=t*a_1 \Rightarrow a_n=(t+1)a_1$ for some $t$ less than $n$. If $t<n-1$, $a_n=0$ or $a_n=a_{t}$, in both cases, contradiction, therefore, $t=n-1$ and $a_n=n*a_1$.
5)Supose that there is an $t<a_2$ such that $t\in B$. Using 3), $a_2-t\in A$, so $a_2-t=a_1 \Rightarrow t=a_1$ We conclude that the minimum value of a $B$ element is $a_1$.
So, $a_1\in B$. If $u\in B$, $u>a_1$ we have that $u-a_1\in A$, using 4), $u-a_1=a_m\Rightarrow u=m*a_1+a_1=(m+1)a_1=a_{m+1}$, for some $m$
Using 5) and 3) We conclude, in this case, $A=B$
6) If the minimum value of a $B$ element was $a_2$, if $u\in B$, $u>a_2$ we have that $u-a_2\in A$, using 4), $u-a_2=a_m\Rightarrow u=m*a_1+a_2=m*a_1+2*a_1=(m+2)a_1=a_{m+2}$, for some $m$
Using 3) y 4), We conclude that $B=\{2a_1, 3a_1, 4a_1, ...\}$.
hectorraul
07.11.2024 13:16
Above solution is incomplete. An infinite set of positive real numbers does not necesarily have a minimum element.
hectorraul
07.11.2024 13:45
$\textbf{Claim 1:}$ $B\subseteq A$ $\text{Proof:}$ First notice that $a\in A$ implies that $ka\in B$ for every positive integer $k$, so $B$ is not bounded above. Then for every $a\in B$ we can find $b, c\in B$ such that $a<b<c$. Now, \[ a<b<c\in B \Rightarrow c-b,b-a\in A \Rightarrow c-a\in B \Rightarrow c-(c-a) = a\in A, \]then $B\subseteq A.$ $\square$ Lets consider two cases: $\textbf{Case 1:}$ $A$ does not have minimum element. For every $c\in A$ we can find $a, b\in A$ such that $a<b<c$. Now, \[ a<b<c\in A \Rightarrow a+b,b+c\in B \Rightarrow c-a\in A \Rightarrow a+(c-a) = c\in B, \]then $A\subseteq B$ and together with the claim we have $A = B$. $\textbf{Case 2:}$ $A$ have a minimum element, say $r$. Clearly $\{2r, 3r,\dots \} \subseteq B $, the claim ensure that $\{r, 2r, 3r,\dots\} \subseteq A $. Suppose that $x\in A$ and $kr< x< (k+1)r$ with $k\ge 1$. Now $kr+x\in B$ and $2kr< kr+x< (2k+1)r$, this means that $A\ni kr+x-(2kr)< (2k+1)r-2kr = r$, contradiction with the minimality or $r$. Then we have that $\{r, 2r, 3r,\dots\} = A$ and $B$ is either $\{2r, 3r,\dots \}$ or $A$, and we are done.
emi3.141592
16.11.2024 15:09
hectorraul wrote: $\textbf{Claim 1:}$ $B\subseteq A$ $\text{Proof:}$ First notice that $a\in A$ implies that $ka\in B$ for every positive integer $k$, so $B$ is not bounded above. Then for every $a\in B$ we can find $b, c\in B$ such that $a<b<c$. It is true that the fact that $a \in A$ implies $ka \in B$, although I do not think it is trivial enough to simply state it. I believe you should delve a bit deeper into the reason why $ka \in B$ so that the proof is complete.
emi3.141592
16.11.2024 15:13
Claim 1.- $A$ and $B$ are not bounded above. Proof.- Clearly, $A$ is bounded above if and only if $B$ is bounded above, so we can assume that both are bounded above. Since $B$ is bounded, for all $\epsilon > 0$, there exist $b_1, b_2 \in B$ such that $|b_1 - b_2| < \epsilon$. It follows that $\inf A = \inf B = 0$, so there exist $a \in A$ and $b \in B$ such that $2a - b \approx 2\sup A - 0$. However, this is a contradiction, since $A$ cannot contain elements greater than $\sup A$. $\square$ Claim 2.- $B \subset A$. Proof.- Since $B$ is not bounded above, for every $b \in B$, there exist $b_1, b_2 \in B$ such that $b < b_1 < b_2$. It follows that $$b = b_2 - ((b_2 - b_1) + (b_1 - b)) \in A$$Therefore, $B \subset A$. $\square$ To conclude, we consider two cases. Case 1.- $A$ does not have a minimum element. Since $A$ has no minimum, for every $a \in A$, there exists $a_1 \in A$ such that $a_1 < a$. It follows that $$a = ((a + a_1) - 2a_1) + a_1 \in B$$Thus, $A \subset B$. Combining this with Claim 2, we obtain that $A = B$. Case 2.- $A$ has a minimum element. Let $m = \min A$. Since $B \subset A$, by an inductive argument, we have that $nm \in A$ for all positive integers $n$, so $\{2m, 3m, 4m, \dots\} \subset B$. Now, if there exists $x \in A$ such that $(k-1)m < x < km$ for some $k \in \mathbb{Z}^+$, then $$2m + x \in B \implies (2m + x) - (k+1)m = x - (k-1)m \in A$$but $x - (k-1)m < m$, which is a contradiction. We conclude that $A = \{m, 2m, 3m, \dots\}$, so $B = \{2m, 3m, 4m, \dots\}$ or $B=A$, and we are done.