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

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?

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 $$f(s) a contradiction. Similarly, $$I$$ cannot contain its right end point.
An open interval is a connected set, and $$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, 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 $$I_m$$.