Let's work through this problem from Vladimir A. Zorich's Mathematical Analysis (Volume I), §2.2 Problem 23 step by step. We're given two real number models $\mathbb{R}$ and $\mathbb{R}'$, plus a map $f:\mathbb{R} \to \mathbb{R}'$ that satisfies:

• Additivity: $f(x+y) = f(x) + f(y)$ for all $x,y \in \mathbb{R}$
• Multiplicativity: $f(x \cdot y) = f(x) \cdot f(y)$ for all $x,y \in \mathbb{R}$

We'll start with the required subclaims, then wrap up by proving $f$ is an order-preserving bijection.

a. Prove $f(0) = 0'$ #

Use the additive property with $x = y = 0$:
$$f(0 + 0) = f(0) + f(0)$$
Simplify the left-hand side, then add the additive inverse of $f(0)$ (denoted $-f(0)$) to both sides in $\mathbb{R}'$:
$$f(0) = f(0) + f(0) \implies 0' = f(0)$$
That's straightforward—$f$ maps the additive identity of $\mathbb{R}$ to the additive identity of $\mathbb{R}'$.

b. Prove $f(1) = 1'$ #

Start with the multiplicative property, setting $x = y = 1$:
$$f(1 \cdot 1) = f(1) \cdot f(1) \implies f(1) = [f(1)]^2$$
Rearrange to get:
$$f(1) \cdot (f(1) - 1') = 0'$$

Since $\mathbb{R}'$ is a real number model, it's a field with no zero divisors—so either $f(1) = 0'$ or $f(1) = 1'$. We can eliminate $f(1) = 0'$ immediately: if this were true, then for any $x \in \mathbb{R}$,
$$f(x) = f(x \cdot 1) = f(x) \cdot f(1) = f(x) \cdot 0' = 0'$$
This would make $f$ the zero map, which can't be a bijection (it's neither injective nor surjective). So we must have $f(1) = 1'$.

Step 1: Prove $f$ is Injective #

Suppose $f(x) = f(y)$ for some $x,y \in \mathbb{R}$. Using additivity, we know $f(-y) = -f(y)$ (set $x = -y$ in $f(x+y) = f(x)+f(y)$ to get $0' = f(-y) + f(y)$). So:
$$f(x - y) = f(x) + f(-y) = f(x) - f(y) = 0'$$

Now assume $x \neq y$, so $x - y \neq 0$. Since $\mathbb{R}$ is a field, $x - y$ has a multiplicative inverse $(x - y)^{-1}$. Using multiplicativity:
$$f((x - y) \cdot (x - y)^{-1}) = f(1) = 1' = f(x - y) \cdot f((x - y)^{-1})$$
But we assumed $f(x - y) = 0'$, which would mean $0' \cdot f((x - y)^{-1}) = 1'$—impossible, since 0' times any element in $\mathbb{R}'$ is 0'. Thus, our assumption is wrong: $x = y$, so $f$ is injective.

Step 2: Prove $f$ is Surjective #

First, note $f$ maps rational numbers to their counterparts in $\mathbb{R}'$:

• For positive integers $n$, $f(n) = f(1+1+\dots+1) = n \cdot f(1) = n'$
• For positive integers $n$, $f(1/n) = (1/n)'$ (since $f(n \cdot 1/n) = f(1) = 1' = n' \cdot f(1/n)$)
• For any rational $q = m/n$, $f(q) = f(m) \cdot f(1/n) = m' \cdot (1/n)' = q'$

Now take any $y' \in \mathbb{R}'$. Since $\mathbb{R}'$ is complete, $y'$ is the limit of some Cauchy sequence of rational elements ${q_n'}$, where each $q_n' = f(q_n)$ for $q_n \in \mathbb{Q}$. The sequence ${q_n}$ is Cauchy in $\mathbb{R}$, so it converges to some $x \in \mathbb{R}$.

Since $f$ preserves order (we'll prove this next), it's continuous with respect to the order topologies of $\mathbb{R}$ and $\mathbb{R}'$. Thus:
$$\lim_{n \to \infty} f(q_n) = f(\lim_{n \to \infty} q_n) = f(x) = y'$$

Alternatively, since $f$ is strictly order-preserving and injective, its image is a dense, unbounded subset of $\mathbb{R}'$. By completeness of $\mathbb{R}'$, this image must equal all of $\mathbb{R}'$—so $f$ is surjective.

Step 3: Prove $f$ is Order-Preserving #

First, show that if $x > 0$ in $\mathbb{R}$, then $f(x) > 0'$ in $\mathbb{R}'$. Since $x > 0$, it has a square root $z \in \mathbb{R}$ such that $x = z^2$. Using multiplicativity:
$$f(x) = f(z^2) = f(z) \cdot f(z) = [f(z)]^2$$

In $\mathbb{R}'$, squares of non-zero elements are positive. Since $z \neq 0$, $f(z) \neq 0'$ (by injectivity, since $f(0) = 0'$), so $[f(z)]^2 > 0'$. Thus, $f(x) > 0'$.

Now suppose $x > y$ in $\mathbb{R}$. Then $x - y > 0$, so $f(x - y) = f(x) - f(y) > 0'$, which implies $f(x) > f(y)$. So $f$ preserves the order relation.

Conclusion #

We've shown $f$ is injective, surjective, and order-preserving. Therefore, $f:\mathbb{R} \to \mathbb{R}'$ is an order-preserving bijection.

内容的提问来源于stack exchange，提问作者fantasie