# how to prove, $f$ is onto if $f$ is continuous and satisfying $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$

let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function such that $|f(x) - f(y)| ≥ |x - y|$ for all $x,y$ in $\mathbb{R}$. then $f$ is onto.

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What have you tried? What theorems about continuous functions do you think might be relevant? Do you know what possible sets can be the range of a continuous function from $\mathbb R$ to $\mathbb R$? Have you noticed any properties of functions satisfying this inequality? For example, is such an $f$ necessarily injective? –  Jonas Meyer Sep 27 '12 at 4:19

Let $A=f(\mathbb R)$. The inequality implies that $f$ is injective with continuous inverse defined on $A$, hence by assumed continuity of $f$, $f:\mathbb R\to A$ is a homeomorphism. The inequality along with continuity of $f$ and completeness of $\mathbb R$ also implies that $A$ is complete with the metric restricted from $\mathbb R$. The only subspace of $\mathbb R$ that is homeomorphic to $\mathbb R$ and complete (with the restricted metric) is $\mathbb R$, so $A=\mathbb R$.

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This seems likely beyond the scope of the course (which is not specified) to me. Maybe it will encourage some to go further. +1 in the hope that it will. –  Ross Millikan Sep 27 '12 at 4:44
Ross: Thank you. I do not intend to match the scope of a course with this answer. –  Jonas Meyer Sep 27 '12 at 4:48

Assume $f$ is not onto, implying that there exist $y\in Y$, here i denote Y as the codomain of $f$, where no $x\in X$, X is the domain of $f$, satisfies $f(x)=y$. Note that $f$ is continuous in the whole real number line. In another words, for any $y\in Y$ as well as in real number line, due to continunity, there should exist and $x$ s.t.$\lim_{t\rightarrow x}f(t)=y$ implies $x$ satisfies $f(x)=y$ for any $y$ in real which is a contradiction

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I don't see an answer. Where is the inequality used? Not all continuous functions from $\mathbb R$ to $\mathbb R$ are onto. –  Jonas Meyer Sep 27 '12 at 4:17
That's no true. $f:\Bbb R\to \Bbb R$ given by $f(x)=e^x$ is continuous in the whole real line, but there is no $x$ such that $\lim_{t\to x} e^x=-1$ for example. –  leo Sep 27 '12 at 4:17

Hint:

The given inequality implies that $f$ must be one-to-one. Without loss of generality, suppose $f$ is increasing.

Suppose now that $f$ does not take the value $M>f(0)$ with $M$ positive. Then by continuity we must have $|f(x)-f(0)|\le M-f(0)$ for all $x\ge 0$. This, however, contradicts your given inequality (consider the inequality with $y=0$ and $x>M-f(0)$). Thus, $f$ is not bounded above.

Now argue in a manner similar to the above that $f$ is not bounded below.

So, you have a continuous function which is neither bounded above nor bounded below...

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If $M$ is connected and $f$ is continuous, then $f(M)$ is connected. $\mathbb{R}$ is connected. Hence $f(\mathbb{R})$ must be an interval. Now you only have to prove that $f(\mathbb{R})$ cannot be bounded from above or below. That's up to you.