I want to be sure about a detail:

We know that, any contractive function is continuous. Reformulated in other words, any function that does not increase distance is continuous.

But, suppose we are in $\mathbb{R}$, and we have a distance increasing function $f$ on a subset $A$ of $\mathbb{R}$. Does the density property of $A$ in $\mathbb{R}$ guarantee the continuity of this function $f$? ie. Will the image of $A$ by $f$ still be dense in $\mathbb{R}$?


No, you can define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = \frac12 x$ for $x \in \mathbb{Q}$ and $f(x) = 2x$ for $x \notin \mathbb{Q}$. $f$ is contractive on the dense set $\mathbb{Q}$ but not on $\mathbb{R}$ and nowhere continuous except at $x=0$.

  • $\begingroup$ you saying there exists non continuous functions which are contracting too, right? $\endgroup$ – freehumorist Jan 4 '19 at 23:49
  • $\begingroup$ @freehumorist No, this function is not contracting and it's not continuous. $\endgroup$ – Henno Brandsma Jan 4 '19 at 23:50
  • $\begingroup$ so how is this an answer to my question? I thought I asked that, when we have a distance-increasing function over a dense set, the continuity would be preserved in the image set too... $\endgroup$ – freehumorist Jan 4 '19 at 23:53
  • $\begingroup$ @freehumorist we also have here a distance increasing function on the dense irrationals and no continuity. So it does answer that too. BTW Distance increasingness does not imply continuity, decreasingness does (but on the whole domain). $\endgroup$ – Henno Brandsma Jan 4 '19 at 23:55
  • $\begingroup$ Ok I got your answer. I think I should add another criteria: I want the function to be linear and monotone. Your example amazed me, now that I get; though it is not the answer I wanted. $\endgroup$ – freehumorist Jan 4 '19 at 23:59

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