Algebraic structure, injection map, standard topology

Question

Suppose that $\varphi : \mathbb{R} \to \mathbb{R}$ is a function which satisfies: $\varphi(1) = 1$ and $$ \varphi(x + y) = \varphi(x) + \varphi(y) , \varphi(x * y) = \varphi(x) * \varphi(y) $$ for all $x, y \in \mathbb{R}$. Notice that we are NOT assuming that $\varphi$ is a map.

a. Show: $\varphi$ is an injection.

b. Prove: $s < t \implies \varphi(s) < \varphi(t)$. Hint: $t - s = u^2$ for some $u \neq 0$.

c. What is $\varphi(\epsilon)$ if $\epsilon \in \mathbb{Q}$?

d. Prove: $\varphi : (\mathbb{R}, \sigma) \to (\mathbb{R}, \sigma)$ is a map. Here $\sigma$ is the standard topology on $\mathbb{R}$.

e. What is $\varphi$? Prove your assertion.

Answer 1

$\varphi$ is the identity. The reason? By b.-, $\varphi$ is an increasing function that (by c.-) fixes $\mathbb{Q}$. Hence, $\varphi$ is a increasing function, continous in a dense subset of $\mathbb{R}$, so $\varphi$ is continous at $x$ for all $x \in \mathbb{R}$ . Finally, an $\epsilon$-$\delta$ argument, aproaching to an irrational number by rational numbers, shows that $\varphi$ should be the identity.

Answer 2

Here is what I have so far. Please correct my errors or ambiguities and help me achieve a complete, detailed proof. If there is a simpler, more elegant way of reaching the solution, please show. Thank you very much for all of your help.

Want to show: $x \neq y \implies \varphi(x) \neq \varphi(y)$

If x = 0, then $\varphi(0) = 0$

a. To show injection for addition: $\varphi(x) + \varphi(y) = \varphi(x + y)$ $\varphi(0 + 0) = \varphi(0) + \varphi (0)$ $\varphi (0) = 0$

To show injection for multiplication: $\varphi (-1) = -1$ $1 = \varphi (-1 * -1) = \varphi (-1)^2 = 1$ $0 = \varphi (-1 + 1) = \varphi(-1) + \varphi(1)$

Since by assumption $\varphi(1) = 1$ So $\varphi(-1)$ must equal -1.

We claim that $\varphi(-y) = \varphi(-1) * \varphi(y) = -\varphi(y)$

Proof by contradiction: Suppose $x \neq y$ but $\varphi(x) = \varphi(y)$ $\varphi(x - y) = \varphi(x) + \varphi(-y) = \varphi(x) - \varphi(y) = 0$ by assumption $\varphi((x - y) * (1/(x-y)) = \varphi(x - y)*\varphi(1/(x-y))$

LHS = $\varphi(1) = 1$ RHS = $0$ Contradiction. Therefore $x = y$

b. From Part (a), $\varphi(x - y) = \varphi(x) - \varphi(y)$

Using the hint, we can show $t - s = u^2$ for some $u \neq 0$ by: $0\leq(\phi(u))^2=\phi(u) * \phi(u)=\phi(u^2)=\phi(t-s)=\phi(t)-\phi(s)$

We know that $\phi(t)=\phi(s)\iff t=s$

However, $t \neq s$ and $\phi(t)-\phi(s) \neq 0$

Since $\phi$ is nonzero and squared, it must be positive

So $\phi(t)-\phi(s) > 0$ and $t > s$.

c. First, we want to show for integers, $\phi(x) = x$

$\phi(0) = 0$

$\phi(2) = \phi(1) + \phi(1) = 2$

$\phi(3) = \phi(1) + \phi(2) = 3$

By induction...

$\phi(x) = x$ and $\phi(-x) = -x$

This holds for rational numbers as well:

$\phi((p/q) * q) = \phi(p/q) * \phi(q)$

$\phi(p) = \phi ((p/q) * q)$

$p = \phi ((p/q) * q)$

$p/q = \phi (p/q)$

Therefore $\phi(\epsilon) = \epsilon$ as long as $\epsilon \in \mathbb{Q}$

$\phi(\epsilon)$ is the identity function on rationals

d. Using an $\epsilon-\delta$ argument: $lim_{x \to c} \phi(x) = \phi(c)$ means for every $\epsilon > 0 , \exists \delta s.t. |x - c| < \delta \implies |\phi(x) - \phi(c)| < \epsilon$

We can choose $\delta$ so that $c + \delta$ is rational.

Pick $a, b \in \mathbb{Q}$

Since $\phi$ is injective by Part (a), positive and increasing by Part (b), order- preserving and fixes $\mathbb{Q}$ by Part (c), we know that $b < c < a$

Take $\delta = min ((b - c), (c - a))$

Then $b \leq c - \delta < c < c + \delta \leq a$

Applying $\phi$, we get the same inequality: $\phi(c) - \epsilon < b < \phi(c) < a < \phi(c) + \epsilon$

Do I need to say anything else to finish the proof for Part (d)?

e. Part (e) is answered by Mro - thank you. Is a more elaborate answer required to fully prove that $\phi$ is the continuous identity on rationals?