# Topology on $\mathbb R$ induced by a strictly monotone function

Let $f: \mathbb R \to \mathbb R$ be a strictly monotone functions. Consider the following metric on $\mathbb R$: $$d(x,y) = |f(x) - f(y)|, \qquad x,y \in \mathbb R.$$ Whether that metric topology is the same as the standard one?

It is the same if $f$ is continuous or has finite number of points of discontinuity. But whether it's the same for arbitrary $f$, e.g. if it discontinuous on $\mathbb Q$?

UPDATE

Actually, I wanted to ask, whether that metric topology always contain the standard one? It is true if there are finite number of points of discontinuity: then we have the standard topology generated by the open intervals plus we have points and semiopen (in usual topology) intervals that are open.

• I believe a strictly monotone function has to be continuous. – Yanko Oct 7 '17 at 22:07
• @yanko $f(x)=x\lfloor 1+x^2\rfloor$ is strictly increasing, yet not continuous. – zwim Oct 7 '17 at 22:11
• @zwim right nice example! – Yanko Oct 7 '17 at 22:13
• @yanko moreover, it can have a countable everywhere dense set of points of discontinuity: math.stackexchange.com/questions/172753/… – vanger Oct 7 '17 at 22:22

Consider the strictly monotonic function $f$ given by $$f(x) = \begin{cases} x & x \le 0\\ x +1 & x > 0\\ \end{cases}$$
It's continuous except at $0$, but in the topology induced by it the sequence $x_n = 1/n$ no longer converges to the point $0$. So your statement about giving the same topology isn't correct.
I haven't proved it, but I think the topology induced by this function is a "separated union" of $(-\infty, 0]$ and $(0, +\infty)$, which you can think of taking regular $\mathbb{R}$ and breaking it at $0$ so that the point $0$ stays attached to the negatives. You can generalize this to multiple points of discontinuity, and think about what the difference would be if the function was right-continuous, or neither left- nor right- continuous, at the points of discontinuity.
• Yeah, of course, I was wrong about the topologies being the same. We could take $f(x) = -1 + x$ for $x < 0$, $f(0) = 0$, $f(x) = 1 + x$ for $x > 0$ and see that $\{0\}$ is open which is not in the usual topology. I wanted to ask, whether that metric topology contains the usual one. Topologies of your and my example contain all sets that open in the usual topology plus some more sets. I updated the question. – vanger Oct 7 '17 at 23:51