Is there an open bijective map from $\mathbb{R}$ to $\mathbb{R}$ that is not continuous?

Question

I came upon this when trying to solve a similar problem first: Open maps which are not continuous(1), which is essentially my problem without requiring the map to be bijective.

To my knowledge, there are a bunch constructions satisfying the weaker constraints: Conway base 13 function(https://en.wikipedia.org/wiki/Conway_base_13_function), a cool one using Riemann Series Theorem (see (1)), and basically all strongly Darboux functions.

The problem is that all these constructions are not bijective, and I'm looking for a bijective example. Immediately this disqualifies all strongly Darboux functions, as they are not bijective on any open set, and this is my progress so far.

Answer 1

The answer to the question in the title is negative.

We know that if a function $f:I \to \Bbb{R}$ is continuous and injective then its inverse $f^{-1}:f(I) \to I$ is also continuous.

Here $I$ denotes an interval .

Now since $f$ is open and a bijection we have that $f^{-1}$ is a continuous function,so $(f^{-1})^{-1}$ is also continuous.