# Division of segments into infinitely many parts.

Let AB and CD be two segments, so that the length of AB is 1, and the length of CD is 2.

If we divide AB and CD in infinitely many parts, how "long" would those parts be? I'm particularly interested in their size, relative to each other.

Intuitively, the length of these parts would approach 0, as we increase their number. But, if it became zero when we have infinitely many segments, then that would be a problem. Because then, by adding up infinitely many parts with length 0, I'd get a 0-length segment back.

So, I'm guessing that the lengths of those parts are not really zero, but some value $\epsilon$, bigger than 0, but smaller than any other real number; the value that comes immediately after zero.

Now that we have this value ϵ, then we could say that the segment AB is composed of infinitely many parts of size ϵ. But, what about CD?

• Would it be composed of twice as many parts of size ϵ?
• Would it still be split into infinitely many parts, but their size being 2ϵ?

While I debated this with a friend, she said that "some infinities are bigger than others", and said that the second segment would be split in twice as many parts as AB, all of size ϵ. To which I replied that if we have infinitely many elements, doubling them doesn't matter, they still are infinitely many.

You could argue that I haven't defined how I split these segments. Let's say that we split them in half, and their halves in half, and so on until infinity. I'm aware that defining the splitting method could make all the difference, but I don't know enough maths to reasonably predict how this would affect the parts. I'm also aware that all I've said above could be seen as complete nonsense: nevertheless, I'm asking it anyway, or I won't be able to sleep tonight without knowing the answer. :P

• Your question of dividing them into "parts" is ambigous. You could break the line into individual points. Then they would each measure 0. Or you could break the into intervals. Every connected interval with more than one point is finite so you would have an infinite collection of finite measures that add to a finite amount. Which is not in the least unusual, paradoxical, or impossible. Or you could break it into sets of diconnected points or any combination. – fleablood Apr 30 '16 at 22:20
• If you divide them into half and half again they will each be finite. 1/2,1/4,1/8,1/16 etc. If you add these up 1/2 + 1/4 + 1/8+.... =1. This is well known. But you are assuming you are dividing it into intervals. You could divide it into {all irrations}, {all integers}, {all rations with denominator 2} ... { with denominator k}. These sets have no length, at least not in the sense you mean. A point has no lengths. Only an uncountable interval of points. Which is confusing, I know. – fleablood Apr 30 '16 at 22:28

## 3 Answers

From the things you say it seems clear that when you talk about splitting $AB$ into infinitely many parts, you mean infinitely many parts of equal length.

You simply can't do that. You seem to realize what the problem is, but you're not drawing the right conclusion: The problem is if the parts have positive length that would mean $AB$ had infinite length, contradiction, while if the parts have zero length that would mean $AB$ had zero length, contradiction.

The conclusion is you simply cannot split $AB$ into infinitely many parts of equal length; asking what would happen if you did is meaningless.

• The parts having zero length does not actually imply the interval has zero length; the Lebesgue measure is countably additive but not uncountably additive. The OP wants to split intervals in half over and over again, and one valid interpretation of that idea is the one I express in my answer which yields uncountably many singleton sets of measure zero. Do you have a different interpretation of completing the process of halving intervals indefinitely? – arctic tern Apr 30 '16 at 21:14
• @arctictern I'm aware that Lebesgue measure is not countably additive. That makes the fact that the segment is in fact an uncountable union of sets of measure zero seem irrelevant. Because what the OP clearly has in mind is a splitting such that the original length is the sum of the lengths of the parts. There's no consistent way to get a splitting with that property out of the fact that the segment is an uncountable union of points, so avoiding that sort of technicality seemed best. It is impossible to split the segment into infinitely many parts of equal length, addtively. – David C. Ullrich Apr 30 '16 at 22:03
• I disagree with your adjective "clearly" and with your implicit assumption that my interpretation is invalid, but desiring a splitting of the interval into equal-length parts such that the original interval's length is the sum of the parts' lengths is at least consistent with what the OP actually wrote, so I now consider that another valid interpretation of OP's question. (Although I'm not sure what it has to do with the halving process.) – arctic tern Apr 30 '16 at 22:38
• @arctictern I didn't say your interpretation was invalid. This is all fuzzy enough there's no such thing as a valid interpretation, imo. The repeated halving thing does seem to me to have been an afterthought on the OP's part. But regarding whether why what seems clear to me seem clear to me, note the "Because then, by adding up infinitely many parts with length 0, I'd get a 0-length segment back.". – David C. Ullrich Apr 30 '16 at 22:46
• If OP were familiar with things like measures and countable additivity, I would agree with your conclusion from that quote, but the OP does not seem very well-versed in these sorts of things. – arctic tern Apr 30 '16 at 23:30

Length is formalized by the mathematical concept of Lebesgue measure, which assumes real number values. A subset of the real number line cannot have infinitessimal measure, it is either zero or positive. Some subsets are nonmeasurable though.

Also, there is no "value that comes immediately after zero." One is free to create their own partially ordered sets, label one of its elements "zero" and by fiat declare there is a thing that comes immediately after it, but then there comes the question of if you're actually talking about numbers or values at all.

As for the number of parts, the way to compare "number of things" is with cardinality. While it is true there are different sizes of infinity (for example there are strictly more real numbers than there are integers), we know that given any infinite cardinal number $\kappa$ the equality $2\kappa=\kappa$ holds (for example there are just as many integers as even integers).

There are many different ways of splitting an interval up into pieces. For instance, we can split the interval $[0,1]$ up into $\{1\}$ (which has measure $0$) plus intervals of lengths $\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots$ or alternatively plus intervals of lengths $\frac{2}{3},\frac{2}{9},\frac{2}{27},\cdots$.

But you've declared your process of splitting up the interval into pieces is just to keep halving them, starting with the original interval. You haven't actually said what it means to complete this process though, so I'll fill in this blank. If $A_1,A_2,A_3,\cdots$ are pieces at stages $1,2,3,\cdots$ of the splitting process respectively, and $A_2$ is broken off from $A_1$, $A_3$ is broken off from $A_2$, etc. then we declare the intersection $\bigcap_{n\ge1}A_n$ to be one of the final pieces "after completion."

Since this is equivalent to specifying the binary representation of a number in the interval $[0,1]$, this means the pieces of the interval $[0,1]$ will just be the singleton sets $\{x\}$ for all $0\le x\le 1$. Thus, starting with any positive-length interval $[a,b]$ in the real number line, there will be $\mathfrak{c}$-many pieces each having Lebesgue measure $0$.

(Note $\mathfrak{c}$ is a cardinal number which we call the "continuum." It is the cardinality of the whole set of reals, and also the cardinality of any positive-length interval $[a,b]$ within the reals.)

In modern mathematics we have several ways of formalizing infinity. The one that is most relevant to your question was provided by Abraham Robinson; see here.

Following his framework, there are both infinitesimals and infinite numbers. Thus if $H$ denotes an infinite number, one can indeed divide a unit interval into $H$ parts. Each part would have length $\frac{1}{H}$, an infinitesimal.

Of course if you wish to divide an interval of length $2$ into infinitesimal subintervals of the same length $\frac{1}{H}$ you would require $2H$ of them rather than just $H$.

This approach to infinity is not to be confused with that of Cantor. Cantorian cardinalities are not useful for answering your question.