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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.

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

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  • $\begingroup$ 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. $\endgroup$ – vanger Oct 7 '17 at 23:51

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