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Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\langle x\rangle >0$. Since $langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4 3 r_x$. It follows that $r_y\leq |y'-y|<r_x$.

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\setminus\langle x\rangle >0$. Since $\langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4 3 r_x$. It follows that $r_y\leq |y'-y|<r_x$.

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\langle x\rangle >0$. Since $langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4/3$. It follows that $r_y\leq |y'-y|<r_x$.

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\langle x\rangle >0$. Since $langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4 3 r_x$. It follows that $r_y\leq |y'-y|<r_x$.

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. Let $\epsilon$ be either $0$ or positive such that $\langle x\rangle$ containing an interval of length $\epsilon$. Let $x'$ be maximal w.r.t. $[x',x'+\epsilon]\subset \langle x\rangle$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. It is sufficient to show that every suffix of $\langle x \rangle$ is an arithmetic sequence. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$. If $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length. By Claim 1, $\langle x \rangle$ must have an empty interior; therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Yet another solution (to the original question).

Define an equivalence relation on $\mathbb R^+$ by $x\sim y$ iff $f(x)=f(y)$. Since $f$ is continuous,

1. every equivalence class is closed.

Let $0<x<y$. Since $y=\frac {x +(2y-x)}{2}$, either $x\sim 2y-x$ or there exists $z\in (x,y)$ such that $z\sim y$. Therefore,

2. for every $x<y$, if [$\forall z\in(x,y)$ $z\not\sim y$] then $x\sim 2y-x$.

It suffices to prove the following proposition.

Proposition 1. Every equivalence relation on $\mathbb R^+$ that satisfies (1) and (2) is trivial (i.e., has one equivalence class).

Proof. Suppose there were a non-trivial equivalence relation "$\sim$" satisfying (1) and (2).

Step 1. Every equivalence class is unbounded and its interior is either unbounded or empty.

Suppose $\langle x\rangle$ were an equivalence class which is either bounded or has a non-empty bounded interior. There exits $\epsilon\geq 0$ such that $\langle x\rangle$ contains an interval of length $\epsilon$, and there exists $x'=\max\{x'': [x'',x''+\epsilon]\subset \langle x\rangle\}$. Take some $y \in (x',\infty)\setminus \langle x\rangle$, and let $y'=\min ( \langle y\rangle\cap[x',\infty))$. By (2), $[x',x'+\epsilon]'\sim 2y'-[x',x'+\epsilon]$ contradicting the maximality of $x'$.

Step 2. Every equivalence class is an arithmetic sequence.

Let $\langle x\rangle$ be an equivalence class. If $x_0=\inf \langle x \langle >0$, then by (2) the elements of $(x_0,2x_0)$ are not in $\langle x\rangle$. Since $\langle x\rangle$ is unbounded and closed, there is a minimal $r_x\geq x_0$ such that $x_0+r_x\in\langle x\rangle$. By (2), the elements of $(x_0,x_0+r_x)$ are not in $\langle x\rangle$ and $x_0+r_x$ is in $\langle x\rangle$. Applying the same argument inductively on all intervals $[x_0+nr_x,x_0+(n+1)r_x]$, we get $\langle x\rangle=\mathbb R^+\cap (x+r_x\mathbb Z)$.

Suppose $\inf \langle x \rangle = 0$ then, the compliment of $\langle x \rangle$ contains an interval $(a,b)$ such that $a,b\in \langle x \rangle$ and starting from that interval $\langle x \rangle$ is an arithmetic sequence with period $b-a$. Since this is the case for any such intervals, they must all have the same length; therefore $\inf \mathbb R^+\langle x\rangle >0$. Since $langle x\rangle$ has a bounded interior it must have an empty interior (by Claim 1); therefore it does not contain a prefix of $\mathbb R^+$, so $\inf \langle x \rangle >0$.

Step 3. For every $x$ and $y$, $r_y \leq r_x$ with strict inequality for some $x,y$. Contradiction!

Let $x\not\sim y$. There exists $x'\sim x$ such that $\langle y\rangle\cap(x',x'+r_x)\neq\emptyset$. By (2), since $\forall z\in(x',x'+r_x)\ z\not\sim x'+r_x$, $\langle y\rangle\cap(x'+r_x,x'+2r_x)\neq\emptyset$. By induction, $\langle y\rangle\cap(x'+nr_x,x'+(n+1)r_x)\neq\emptyset$, for all $n\in\mathbb N$; therefore $r_y\leq r_x$. Take $x$ and $y=x+\frac 2 3 r_x$. By (2), $y\sim y'=x+\frac 4/3$. It follows that $r_y\leq |y'-y|<r_x$.