I was trying to understand this answer here but got stuck. It's clear to me that $\varphi: \mathbb R \to \mathbb R$ should map positive numbers to positive numbers and that it follows from that that $\varphi$ is increasing.
But I can't prove that this implies that $\varphi$ is continuous.
How can I prove that if $\varphi: \mathbb R \to \mathbb R$ is increasing and bijective then it is continuous?
If it is increasing but not continuous, then it must have jump discontinuity, i.e. left limit of $\phi$ at a point $a$ is strictly less than the right limit of $\phi$ at $a$. In short, $\phi(a^-)<\phi(a^+)$. Then we can use increasing property to check that $\phi$ can never take value in the set $(\phi(a^-),\phi(a^+))-\{\phi(a)\}$ (which is nonempty since $(\phi(a^-),\phi(a^+))$ is uncountable), contradiction to surjectivity.
First show that if $\sigma : \mathbb R \to \mathbb R$ is an automosphism, then $\sigma(r)=r, \forall r \in \mathbb Q.$ Then using the fact that $\sigma$ is increasing, show that $-\frac{1}{m} < a-b < \frac{1}{m} \Rightarrow -\frac{1}{m} < \sigma (a) - \sigma (b) < \frac{1}{m}, \forall m \in \mathbb N.$
I think increasing is enough to proof, that f is continuous. (I would like to have a 2th opinion to prevent a made a hasty mistake)
Let $ f(x_1) = a $ and $ f(x_2) = b $ for $ x_1 < x_2 $, and $ a < b < \infty $.
Then you can pick an increasing sequence of infinity many points $ x_i $in $ [x_1, x_2] $. If this sequence has a limit point, then it converges and thus must be continuous at that point. Therefore the sequence must not have a limit point, so it does not converges and therefore there is a $\epsilon > 0 $ : $ \forall i,j \in N $ : $ | x_i - x_j | \ge \epsilon $, $i \neq j ;$
But then f is not bound in that given interval, and there is surely a $ x_k $ in your sequence with: $ f(x_k) > b $, which contradicts that f is increasing. Keep in mind that increasing also implicates, that f is injective.
