Prove that $f(x+y)=f(x)+f(y)$, and $f(x\cdot y) = f(x)\cdot f(y)$ implies $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves order.

Question

Q: Prove that $f(x+y)=f(x)+f(y)$, and $f(x\cdot y) = f(x)\cdot f(y)$ implies $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves order.

Original question ($\S$2.2/23 in Mathematical Analysis I by Vladimir A. Zorich)

• Show that if $\mathbb{R}$ and $\mathbb{R}'$ are two models of the set of real numbers and $f:\mathbb{R} \to \mathbb{R}'$ is a mapping such that $f(x+y)=f(x)+f(y)$ and $f(x\cdot y) = f(x)\cdot f(y)$ for any $x, y\in\mathbb{R}$, then

  a. $f(0) = 0'$

  b. $f(1) = 1'$ if $f(x) \not\equiv 0'$, which we shall henceforth assume;

  c. $f(m) = m'$ where $m\in \mathbb{Z}$ and $m'\in \mathbb{Z}'$, and the mapping $f:\mathbb{Z} \to \mathbb{Z}'$ is bijective and preserves the order.

  d. $f(\frac m n) = \frac m n$, where $m, n \in \mathbb Z$, $n\neq0$, $f(m) = m'$, $f(n)=n'$. Thus $f:\mathbb{Q} \to \mathbb{Q}'$ is a bijection and preserves the order.

  e. $f:\mathbb{R} \to \mathbb{R}'$ is a bijective mapping that preserves the order.

$^\ast$By saying '$\mathbb{R}'$ is a model of the real numbers', it means that $(\mathbb{R}', +, \cdot, \le)$ satisfies the axioms of real numbers.

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I think that I could finish the proof of a., b., c. and d., however I failed to prove e.

Actually, I have found this question while searching. However, this is the next question in that textbook, and also there isn't an answer.

Using the result of d., which says $f:\mathbb{Q} \to \mathbb{Q}'$ is a bijection and preserves the order, I tried to demonstrate the real numbers using least upper bounds. For example, suppose $S\subset\mathbb{Q}$ has an upper bound, then $\exists r\in\mathbb{R}$ such that $r = \sup S$. It is also possible to prove that the image of $S$ has an upper bound, therefore $\exists r'\in\mathbb{R}'(r' = \sup f(S))$. However, I don't know how to prove that $r$ and $r'$ are equal.

As I realized that there could be a way to tell that $f$ is a bijection if $f$ preserves the order, I tried to find a contradiction assuming $\exists c \in\mathbb{R}$ where $c<0 \wedge f(c)>0$. However, I still fail to find anything so far.

Could you help me? Thanks.

Answer 1

Note that if $x \geq 0$, then $f(x) = f(\sqrt{x} \cdot \sqrt{x}) = f(\sqrt{x}) \cdot f(\sqrt{x}) = (f(\sqrt{x}))^{2} \geq 0'$, since the axioms of the reals imply that the square of any element is nonnegative, and any nonnegative element has a square root.

Now, if $a \geq b$, $a - b \geq 0$, and by the above $f(a - b) \geq 0$, so $f(a) - f(b) \geq 0$, so $f(a) \geq f(b)$.

Now we show that if $a > b$, we cannot have $f(a) = f(b)$.

If $x > 0$, by the density of the rationals there exists a rational number $q$ with $0 < q < x$. By part d, $0' < f(q)$, and we know $f(q) \leq f(x)$, so $f(0) < f(x)$. This implies if $a > b$, $f(a) > f(b)$. So $f$ is injective and order-preserving.

The last step is to show that $f$ is surjective, for which you'll have to use the completeness property of $\mathbb{R}$ and the fact that $\mathbb{Q}'$ is dense in $\mathbb{R}'$.

Answer 2

(This is an extend comment in response to @virtualize.)

I hope the following ideas could bring you or someone an insight.

Assume $\mathbb{R} = \mathbb{R}'$.

Lemma. If $f : \mathbb{R} \to \mathbb{R}$ is a non-linear additive function, then its graph $ G = \{(x, f(x)) : x \in \mathbb{R}\} $ is dense in $\mathbb{R}^2$.

Proof. Start with any nonzero $x_1 \in \mathbb{R}$. So there exists nonzero $x_2 \in \mathbb{R}$ such that $$ \frac{f(x_1)}{x_1} \neq \frac{f(x_2)}{x_2} $$

otherwise $$ f(x_2) = \dfrac{f(x_1)}{x_1} x_2 $$ would be linear.

This implies $$ \det \left( \begin{bmatrix} x_1 & f(x_1) \\ x_2 & f(x_2) \end{bmatrix} \right) \neq 0 $$

Hence $v_1 = (x_1, f(x_1))$ and $v_2 = (x_2, f(x_2))$ forms a basis of $\mathbb{R}^2$. That is, for all $v \in \mathbb{R}^2$ exists $(r_1, r_2) \in \mathbb{R}^2$ such that $v = r_1 v_1 + r_2 v_2$.

Since $\mathbb{Q}^2$ is dense in $\mathbb{R}^2$, every $v \in \mathbb{R}^2$ can be arbitraly approximated by some appropriately chosen $(q_1, q_2) \in \mathbb{Q}^2$ as $$ v = q_1 v_1 + q_2 v_2 = (q_1 x_1 + q_2 x_2,\ f(q_1 x_1 + q_2 x_2)) $$

Thus $$ G_\mathbb{Q} = \{ (q_1 x_1 + q_2 x_2,\ f(q_1 x_1 + q_2 x_2)): (q_1, q_2) \in \mathbb{Q}^2 \} $$ is dense in $\mathbb{R}^2$. Therefore $G$ is dense in $\mathbb{R}^2$ as $G_\mathbb{Q} \subset G$.

Theorem. An additive and multiplicative function $f : \mathbb{R} \to \mathbb{R}$ is linear.

Proof. For any non-negative $x$ we have $ f(x) = f(\sqrt{x}\sqrt{x}) = f(\sqrt{x})^2 \ge 0 $. Since $f$ avoids the set $\{(x,y) \in \mathbb{R}^2 : x < 0, y < 0 \}$ it cannot be dense in $\mathbb{R}^2$. Hence it follows from from lemma that $f$ is linear. Thus $f(x) = c x$ for some $c \in \mathbb{R}$.

Furthermore, since $ c x y = f(xy) = f(x) f(y) = c x\ c y = c^2 x y$ it follows that $c = c^2$. Then $f(x) = 0$ or $f(x) = x$.

Due to this I heavily suspect the function $f : \mathbb{R} \to \mathbb{R}'$ is precisely the identity between the two models, $f(x) = x'$. Perhaps one could adapt this proof to different models.

Answer 3

$f(x^{2})=(f(x))^{2} \geq 0$ so $f(t)=f((\sqrt t))^{2}) \geq0$ for any $t\geq 0$. Now $x<y$ implies $f(y)-f(x)=f(y-x) \geq0$. Hence f is increasing. Together with d) this gives e).