A question in answer of a user in the question every bijection $f:\mathbb{R}\to[0,\infty)$ has infinitely many discontinuity

Question

This particular question :

Show that every bijection $ f:\mathbb{R} \to [0,\infty)$ has infinitely many points of discontinuity.

was asked in a quiz of mine.

Unable to solve it, I searched on MSE. I found this particular solution.

Points of discontinuity of a bijective function $f:\mathbb{R} \to [0,\infty)$

But I have a question in solution. But both the asker and answerer are not seen on the website for a very long time.

So I am asking my doubt as a separate question :

In the third line of answer given in above link how does author deduce that $f(I_m)$ is an open interval? It means that $f$ maps open intervals to open intervals? Why?

Can anyone please give a rigorous answer?

Answer 1

If $f$ is continuous and injective on an open interval $(a,b)$ then $f$ is monotonic. Suppose $f$ is increasing. By IVP of continuous functions the image is an interval, call it $I$. Suppose this interval contains one of its end points. Say $I=[t,s)$. Then $t=f(x)$ for some $x \in (a,b)$. Pick any $s$ between $a$ and $x$. Then $f(s) <f(x)=t$ a contradiction. Similarly, $I$ cannot contain its right end point.

Answer 2

An open interval is a connected set, and $f$ is continuous, so $f[I_m]$ is connected. The only connected subsets of the real line are intervals (open, half-open, or closed), rays (open or closed), and $\Bbb R$ itself, so $f[I_m]$. If you’re not familiar with the general topological notion of connectedness, you can use the intermediate value theorem to show that $f[I_m]$ must be of one of those types. The crucial point is that these are the convex subsets of $\Bbb R$: if $x$ and $y$ are members of one of these sets, and $x<z<y$, then $z$ is also a member of that set.

As is pointed out in the proof, $f\upharpoonright I_m$, being continuous and injective, is (strictly) monotone, so it is either strictly order-preserving or strictly order-reversing. Since $I_m$ is an open interval or open ray, this means that $f[I_m]$ must also be an open interval or open ray: if it had an endpoint, that endpoint would have to be the image of an endpoint of $I_m$.