Let be $I, J$ any intervals of the real line and $f: I \longrightarrow J$ a bijection such that $x < y \Longrightarrow f(x) < f(y)$. Prove that $f$ is continuous.
I would like to know if my attempt is correct. Thanks in advance!
My attempt:
We show that $f$ is continuous in $x_0 \in I$. Suppose that $f$ is descontinuous in $x_0 \in I$, this is, $\exists \epsilon > 0$, $\forall \delta > 0;$ $|x – x_0| < \delta$ e $|f(x) – f(x_0)| \geq \epsilon$. Let's look at two cases:
(1) $f(x) \geq f(x_0) + \epsilon$:
As $J$ is an interval, exists $a \in (x_0 – \delta,x_0 + \delta)$ such that $f(a) = f(x_0) + \epsilon$ and exists $c \in J$ tal que $f(x_0) < c < f(a)$, so exists $z \in I$ such that $f(z) = c$ and follows from the fact that $f$ is increasing that $x_0 < z < a$, so $z \in (x_0 – \delta,x_0 + \delta)$, which contradicts the discontinuity of $f$ in $x_0 \in I$.
(2) $f(x) \leq f(x_0) – \epsilon$:
As $J$ is an interval, exists $a \in (x_0 – \delta,x_0 + \delta)$ such that $f(a) = f(x_0) – \epsilon$ and exists $c \in J$ such that $f(a) < c < f(x_0)$, so exists $z \in I$ tal que $f(z) = c$ and follows from the fact that $f$ is increasing that $a < z < x_0$, so $z \in (x_0 – \delta,x_0 + \delta)$, which contradicts the discontinuity of $f$ in $x_0 \in I$.
Therefore $f$ is continuous in $x_0 \in I$.
Best Answer
I would show continuity as follows:
Fix $x_0\in I$ and show that $f$ is continuous at $x_0$.
Let $\epsilon>0$, and we may assume $\epsilon$ small enough so that $[f(x_0)-\epsilon,f(x_0)+\epsilon]\subseteq J$.
Since $f$ is a bijection, there are elements $a,b$ such that $f(a)=f(x_0)-\epsilon$ and $f(b)=f(x_0)+\epsilon$, with $a,b\neq x_0$ (here I am using surjectivity and injectivity of $f$)
Now, since $f$ is increasing and $f(a)=f(x_0)-\epsilon < f(x_0)$, we must have $a<x_0$. Similarly, $x_0<b$.
Let $\delta=\min\{x_0-a,b-x_0\}>0$. Then, for any $x\in (x_0-\delta,x_0+\delta)\subseteq (a,b)$, we have
$$f(a)=f(x_0)-\epsilon <f(x)<f(x_0)+\epsilon=f(b).$$
So, $f$ is continuous at $x_0$.