Let us define a "scattered" function as a function f with the following property: For any a, there exists an positive number e such that for any a' not equal to a such that |a'-a|=e. In other words, when we consider the graph (x,f(x)), there is a square centered on (x,f(x)) which contains no other points.

The question is, can we construct a scattered defined on all reals? If not, is it possible to construct a scattered function defined on some uncountable subset of the reals?

P.S. If an actual name exists for what I called a scattered function, I would like to hear it.


I'm not clear on your question, but I think that you may be trying to say that every point on the graph of $f$ is isolated in the plane. Is that right? In this case, we can associate to each point $x$ in the domain of $f$ an open ball inside your square containing $(x,f(x))$ and no other points of the graph, such that the ball has rational center and rational radius Since no two points on the graph can be associated with the same such rational ball, this means that our association is one-to-one. But since there are only countably many rational numbers, the number of such rational balls is countable, and so the domain of $f$ must be countable. Thus, we cannot do it with an uncountable domain.

What the argument shows is that if a subset of the plane has only isolated points, then it is countable.

Another possible interpretation of your question, based on what you wrote, is that you meant to speak just of the boundary of the square, rather than it's interior. In this case, it is possible to do it with uncountable domain. Let $V$ be a Vitali set, which contains exactly one real number in each equivalence class, where $x$ and $y$ are equivalent if $x-y$ is rational. Define $f(x)=x$ for $x\in V$. Now consider any $x\in V$. Note that $x\pm q$ is not in $V$ for any rational number $q\in\mathbb{Q}$. So the square centered at $(x,f(x))$ with side length $2q$ stretches from $x-q$ to $x+q$ both horizontally and vertically, and contains no points from the graph of $f$.

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  • $\begingroup$ Yes, you understood the question correctly. Also, nice answer. $\endgroup$ – yrudoy Dec 21 '10 at 23:26
  • $\begingroup$ I've realized that with the second interpretation (which you've indicated was not your intention), you can actually attain a total function $f:\mathbb{R}\to\mathbb{R}$. Simply let $V$ be the Vitali set, and then define $f(x+q_n)=x+n$, where $x\in V$ and $q_n$ is the $n$-th rational number. This amounts to countably many copies of the earlier example, separated at unit distances. This function is now total, but still exhibits the square-boundary-omitting requirement as before, since the argument works separately on each piece, using a square with rational side less than $1$. $\endgroup$ – Joel David Hamkins Dec 22 '10 at 4:04
  • $\begingroup$ In my previous comment, I meant $f(x+q_n)=x+q_n+n$, so that the graph lies on a shift of the identity function. $\endgroup$ – Joel David Hamkins Dec 22 '10 at 11:25

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