# Show that there exists no function $f:\mathbb{R} \to \mathbb{R}$ such that $|f(a)-f(b)| \geq 1$ $\forall a,b \in \mathbb{R}$ where $a\neq b$

Show that $$\nexists$$ $$f:\mathbb{R} \to \mathbb{R}$$ such that $$|f(a)-f(b)| \geq 1$$ $$\forall ~ a,b \in \mathbb{R}$$ where $$a\neq b$$

I have no idea on how to do this. I think I have an idea to show that we can/cannot (I'm not sure if we can) construct a function when $$f:[p,q]\to \mathbb{R}$$ for some real numbers $$p,q$$. In that case, if we consider the real axis of the interval $$[p,q]$$ and bend it to form a circle, we can perform an inversion about another circle $$C$$ centered at $$O$$ such that our former circle passes through the center $$O$$. Each point on the circle maps to a distinct new point on the line (formed due to the inversion). Let that be the new real axis.

Each point on the real axis is mapped, is there any way that the points might be separated by at least a unit?

• Take $a=b$ to get a contradiction! – Kavi Rama Murthy Jul 26 '19 at 8:40
• your thoughts are way too crazy. note that, for each $n \in \mathbb{Z}$, there is at most one $a \in \mathbb{R}$ for which $f(a) \in [n,n+1]$, so we naturally get an injection from $\mathbb{R}$ into $\mathbb{Z}$, but there is no such thing. – mathworker21 Jul 26 '19 at 8:44
• It should at least be “for all $a,b\in\mathbb{R}$ with $a\ne b$”. – egreg Jul 26 '19 at 8:48
• @Mathejunior no. if you have a specific question and/or point out a specific part you don't understand, I'll be able to help. I made several claims in my comment; be specific about what you don't get. – mathworker21 Jul 26 '19 at 9:01
• it's a well known fact that $\mathbb{R}$ is uncountable. Look that up. – mathworker21 Jul 26 '19 at 9:11

We expand mathworker21's comment. If such function $$f$$ exists then $$\mathbb{R}=f^{-1}(\mathbb{R})=f^{-1}\left(\bigcup_{n\in\mathbb{Z}}[n,n+1)\right)=\bigcup_{n\in\mathbb{Z}}f^{-1}([n,n+1)).$$ Now note that $$|f(a)-f(b)| \geq 1$$ $$\forall a,b \in \mathbb{R}$$ with $$a\neq b$$ implies that the cardinality of $$f^{-1}([n,n+1))$$ is $$0$$ or $$1$$. Hence the set $$f^{-1}(\mathbb{R})$$ is countable. On the other hand, $$\mathbb{R}$$ is uncountable. Contradiction!