# Existence of metric $d$ on $\mathbb R$ such that the function $f:(\mathbb R,d) \to (\mathbb R,d)$ ; $f(x)=-x$ is everywhere discontinuous

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.

• What about the topology generated by the intervals $[a,b)$ ? – user171326 Jun 15 '16 at 8:29
• @N.H. : Sorry but I have edited my question , I want only metric now – user228169 Jun 15 '16 at 8:32
• 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 8:33
• @Crostul : Yeah , from the accepted answer to this question math.stackexchange.com/questions/1825351/… 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 – user228169 Jun 15 '16 at 8:36
• @Crostul : I think , for your metric , the function $f(x)=-x$ "is" continuous at every rational number ... – user228169 Jun 15 '16 at 9:12

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.
• 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 $\phi \circ f \circ \phi^{-1}$ ... ? – user228169 Jun 16 '16 at 11:20
• @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