real analysis - Proving that a monotone additive function is continuous on $Bbb R$

The following is from Bartle's Elements of Real Analysis.

A function on $\Bbb R$ is additive if $f(x + y) = f(x) + f(y)$ for
each $x, y \in \Bbb R$. Prove $(a)$ An additive function that is
continuous at $x = 0$ is continuous everywhere and $(b)$ A monotone
additive function is continuous on $\Bbb R$.

I proved $(a)$. This is a sketch of what I came up with for $(b)$.

$$f(0) = f( 0 + 0 ) = f(0) + f(0) \implies f(0) = 0$$

$$0 = f(0) = f(x - x) = f(x) + f(-x) \implies \forall x \in \Bbb R \; f(-x) = -f(x)$$

For $n \in \Bbb N$ we have $f(nx) = f(x) + ... + f(x) = nf(x)$ and $n f(\frac{x}{n}) = f( \frac{nx}{n}) = f(x)$ which implies that $f(\frac{x}{n}) = \frac 1 n f(x)$. We also have that $a \gt 0 \implies f(a) \ge 0$.

So from these properties we can conclude that $f(q) = q \cdot f(1)$ for any rational number $q$.

Now suppose $(x_n)$ is any sequence that tends to $0$. Then we can find a rational sequence $(q_n)$ - by looking in $(x_n - \frac 1 n, x_n + \frac 1 n)$ - which tends to $0$ such that, $$0 \le |x_n| \le |q_n|$$

Whence by the monotony of the function , $$\forall n \in \Bbb N \;\;\;\; 0 \le f(|x_n|) \le f(|q_n|) = |q_n| \cdot f(1) \implies f(|x_n|) \to 0$$

Now we can show by considering cases that $|f(x_n)| = f(|x_n|)$ and hence we have that $f(x_n) \to 0$. So $f$ is continuous at $0$ and hence from part $(a)$ we can conclude that it is continuous on $\Bbb R$.

I have a couple of questions:

1. Is my solution correct? Are there any unauthorised assumptions in it?



2. There has to be a prettier way of solving this. Can someone please give me a hint?

Sorry for the long prose. Thanks for taking the time to read and for responding.

Answer

Your proof seems fine. For a nicer way:

Assume wlog that $f$ is increasing.

Suppose $f$ is not continuous at $0$. Then there exists $\epsilon > 0$ such that for all $\delta > 0$ there is $0 < x < \delta$ with $f(x) > \epsilon$. (This uses $f(-x) = -f(x)$ to force $0 < x < \delta$). But this implies $f(y) > \epsilon$ for all $y > x$, and so $f(y) > \epsilon$ for all $y > 0$.

But $f(y) = f(y/2 + y/2) = f(y/2) + f(y/2) > \epsilon + \epsilon = 2\epsilon$. So $f(y) > 2\epsilon$ for all $y > 0$. Repeating this, we get $f(y) > 2^n \epsilon$ for all $y > 0$ and $n \in \mathbb{N}$. But this is clearly impossible.

So $f$ is continuous at $0$, and so by part (a) $f$ is continuous everywhere.