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Does there exist a metric $d$ on $\mathbb R$ such that the function $f:(\mathbb R,d) \to (\mathbb R,d)$ defined as $f(x)=-x$ is everywhere discontinuous ?

It is motivated from this question which concerns discontinuity at atleast one point only. Apparently, when we want the function to be discontinuous everywhere, the metric got from a permutation over the real line doesn't seem to be working as observed in the comments.

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  • What about the topology generated by the intervals $[a,b)$ ? –  Jun 15 '16 at 08:29
  • @N.H. : Sorry but I have edited my question , I want only metric now –  Jun 15 '16 at 08:32
  • Take a very bad permutation of the reals $\phi: \Bbb{R} \to \Bbb{R}$ and call $d(x,y)= |\phi(x)- \phi(y)|$. This will give you an ugly metric, for which many standard continuous maps won't be continuous in this metric. I suggest $$\phi (x) = \begin{cases} x & \mbox{ if } x \in \Bbb{Q} \ -x &\mbox{ otherwise} \end{cases}$$ – Crostul Jun 15 '16 at 08:33
  • @Crostul : Yeah , from the accepted answer to this question http://math.stackexchange.com/questions/1825351/does-there-exist-a-metric-d-on-mathbb-r-such-that-the-map-f-mathbb-r-d I know such technique works when I want at least one discontinuity point , but I honestly cannot figure out what to do If every point were to be a discontinuity point . Could you please elaborate –  Jun 15 '16 at 08:36
  • @Crostul : I think , for your metric , the function $f(x)=-x$ "is" continuous at every rational number ... –  Jun 15 '16 at 09:12
  • I know. We should work with something similar. – Crostul Jun 15 '16 at 09:45
  • @Crostul : I tried $g(x)=1/x$ for irrational $x$ but that also doesn't seem to be working :( –  Jun 15 '16 at 12:34
  • may the close voter please explain why this question needs to be closed ? Is it not an interesting question or is it too trivial ? –  Jun 15 '16 at 12:35
  • @user228169 You can probably get this question reopened if you link the other Q, and explain why your attempts mentioned in the comments don't seem to work, in an edit to the Q. Right now it's just a bare problem statement, which is usually discouraged and quickly closed as "missing context" on M.SE. – Mario Carneiro Jun 15 '16 at 19:14

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Consider $$\phi(x)=\begin{cases}x+1&x\in\Bbb Q\\x&x\notin\Bbb Q\end{cases}$$ and define the metric $d(x,y)=|\phi(x)-\phi(y)|$. For any $a\in\Bbb R$ and $\epsilon>0$, we find $y\in\Bbb R$ such that $|a-y|<\epsilon$ and exactly one of $a,y$ is rational and one is irrational. If $a$ is rational, let $x=y+1$, so $d(a,x)=|a+1-x|=|a-y|<\epsilon$; and if $a$ is irrational, let $x=y-1$, so $d(a,x)=|a-(x+1)|=|a-y|<\epsilon$. Then $$d(-a,-x)=|-a+1-(-x)|=|-a+1+y+1|\ge 2-|a-y|>2-\epsilon$$ for rational $a$ and $$d(-a,-x)=|-a-(-x+1)|=|-a+y-2|\ge 2-|a-y|>2-\epsilon$$ for irrational $a$.

In simple terms: We make our open sets look like usual, but with all rational points pushed aside in one direction. The map $x\mapsto -x$ then pushes the rational points into the "wrong" direction.

Hagen von Eitzen
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    Very nice, this was the same example I was thinking of. For comparison to the other Q, one can calculate the $\phi$-image (group conjugate) of $f$, $$\phi^{-1}\circ f\circ\phi(x)=\begin{cases}-x-2&x\in\Bbb Q\-x&x\notin\Bbb Q\end{cases},$$ and then observe that this is not continuous anywhere in the standard metric, to avoid the $\delta$-$\epsilon$ arguments. – Mario Carneiro Jun 16 '16 at 10:28
  • @MarioCarneiro : actually $\phi :(\mathbb R,d) \to \mathbb R$ is continuous , so don't you actually wanna take $\phi \circ f \circ \phi^{-1}$ ... ? –  Jun 16 '16 at 11:20
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    @user228169 Oops, my mistake, yes it should be the other way around. Of course, $$\phi\circ f\circ\phi^{-1}(x)=\begin{cases}-x+2&x\in\Bbb Q\-x&x\notin\Bbb Q\end{cases}$$ is not continuous anywhere either. – Mario Carneiro Jun 16 '16 at 11:30