I Divisibility of bounded interval of reals

[Jarvis323]

Can (0,1)\subset\mathbb{R} be divided into an infinite set S of non-empty disjoint subsets? It seams like any pair of points in different subsets of the partitioning must have a finite difference, and so there must be some smallest finite difference overall, d where |S| \leq 1/d. Can someone point me to the result of this question?

 

[fresh_42]

Without any further requirements, the following trivial example does the job:
$$(0,1) = \bigcup_{x \in S = (0,1)} \{x\}$$

 

[mathman]

There doesn't have to be a a smallest interval. (0,1/2),(1/2,3/4),(3/4,7/8), etc. is a disjoint infinite set of intervals each of length > 0, which fill up the unit interval.

 

[Jarvis323]

I see, yes the obvious solutions are trivial, but still the intuition is difficult. The resolution seams that there is no smallest difference, but also there are no infinitely small differences.

I was trying to consider what the implications of infinite divisibility of space are for the types of structures that are theoretically possible for dynamical systems. I was imagining a finite region in an N dimensional state space, which is occupied by infinitely many N-1 dimensional disjoint layers that are alternating sections of basin of attraction for two separate attractors.

Could you break the unit interval into infinitely many ordered disjoint intervals, such that only every other one has a property P? It seams you can have adjacent intervals, such as (0,1)[1,2), but the adjacent point from (0,1) isn't 0.999..., and it also cannot be 0.999...y..., y <9, and it cannot have a finite number of digits, otherwise there would be another number closer to 1. So it would seam that the concept of adjacency of real numbers is paradoxical? If there are two adjacent points, one of the points cannot be infinite or finite in number of digits, so such a number does not exist, there are only infinitely many that approach being in the other interval but never succeed. It seams like if we limit them or don't, then the intervals are not disjoint or there are missing numbers.

I guess I am just revisiting Zeno's paradox. The logical conclusion seams that the real line cannot be divided at all! At least, it seams we cannot color every other real number a different color. After further reading, I guess it's just well ordering of the reals that I am wondering about. It seams that one does not exist for \leq.

 

[mathman]

There is no such thing as adjacent real numbers, or adjacent rational numbers. Why does this prevent the real numbers from being divided?

Zeno's paradox is resolved by taking into account time, as well as a finite sum for an infinite series.

 

[Jarvis323]

I’m not claiming they can’t be, just saying it seams like they cannot be, because there is no way to separate the left and right parts at any given point, for the same reason you cannot well order them by magnitude. Wherever you divide them there is some kind of mysterious paradoxical fuzz that cannot be cleared up, without resulting in the subsets not being disjoint, or there unions having holes.

But anyway ignoring other details, we can say that a finite region in a real state space can have only finitely many sections of basin of attraction? Or (I think) equivalently we cannot 2 color partitions of the unit interval?

 

[mathman]

Wherever you divide them there is some kind of mysterious paradoxical fuzz that cannot be cleared up, without resulting in the subsets not being disjoint, or there unions having holes.

What is this fuzz? When you divide an interval, you define one division point and it can be included in either subset, but not both, when yoy want them disjoint.

 

[Mark44]

But anyway ignoring other details, we can say that a finite region in a real state space can have only finitely many sections of basin of attraction? Or (I think) equivalently we cannot 2 color partitions of the unit interval?

I'm not sure I understand what you're asking, but I'll take a stab anyway. It's probably easier to comprehend working with the interval [0, 1] than an n-dimensional space. Wouldn't each interval in the example given earlier, the partition $\{(0, 1/2), (1/2, 3/4), (3/4, 7/8), \dots, (\frac{2^n - 1}{2^n}, \frac{2^{n+1} - 1}{2^{n + 1}}), \dots \}$, serve as basins of attraction? If the goal is to color the interval in two colors, you could omit every other interval of the partition above.

 

[Jarvis323]

It would seam like that works, I guess with a countably infinite partition it should work in general, but not uncountable? As long as none of the intervals have zero width, and they can be seqentially ordered, it should be fine I guess.