Dividing the continuum and adding recurring decimals

Question

Let us separate all of the positive numbers contained within the integer "1" into two congruent sets:

[0,0.5), [0.5,1)

In this case, the value "1" is excluded from the sets, we can consider this to be an infinitesimal remainder created by the division. The sets contain 49.999r% of the continuum respectively.

Now suppose that we separate the continuum "1" into two sets that are unequal by an infinitesimal amount, as so:

[0,0.5),[0.5,1]

In this case, the entire continuum is covered, and the sets contain 49.999r% and 50% respectively.

My question is this, if the second case covers more of the continuum than the first case, what total percentages of the continuum do each of the cases?

In other words, is 49.999r + 49.999r = 49.999r + 50?

It would seem to me that 49.999r + 49.999r is two-infinitesimals less than 100, while 49.999r + 50 is simply 99.999r?

Is or is this not the case, and why?

[Please do not answer '0.999r = 1' as this would not be taking the question seriously]

Answer 1

Since we're talking about length, note that for any closed interval $[a,b]$ with $a\le b,$ we have that the length of $[a,b]$ is $b-a.$ as an immediate consequence, the length of any singleton $\{b\}=[b,b]$ is $0.$ Since we want length to be additive--that is, if $A$ and $B$ are disjoint sets with a defined length, we want the length of $A\cup B$ to be the sum of the lengths of $A$ and $B$--it then follows just as immediately given an interval $[a,b)$ with $a\le b,$ its length must be $b-a.$ Thus, all of the subintervals in question have length $\frac12.$

For more detail on this, look into Borel measure and Lebesgue measure.

Answer 2

This collection of comments is too long for a couple comments and might answer your question.

Comments on Terminology

all of the positive numbers contained within the integer "1"

$0$ is not usually taken to be positive. I think you meant "non-negative".

More importantly, "contained within" generally has one of two meanings in math: a subset or an element. But the integer $1$ isn't usually thought of as a set. I think you meant the interval $[0,1]$ in some number system. For now I'm being agnostic about which number system, but the convention is overwhelmingly for interval notation to mean real numbers, unless otherwise specified.

...we can consider this to be an infinitesimal remainder created by the division

Just to be clear, if there is a remainder, then we didn't separate "all" of the numbers into the two sets. Rather, we separated most of them. This made things a little harder to read for me.

More importantly, you have used the word "infinitesimal", perhaps with an intent to describe the length of $[1,1]$. Traditionally, the length of that set would be $|1-1|=0$.

Suppose we use a number system (with order and arithmetic operations) other than the reals so that infinitesimal values can exist. If the length (meaning $|b-a|$) of some interval $[a,b]$ or $[a,b)$ is some infinitesimal value $\varepsilon$, then (if $a\le b$) I would expect a number like $a+\varepsilon/2$ to fit inside the interval between $a$ and $b$. Nothing like that can be done in $[1,1]$ in any number system: the only number between $1$ and $1$ (inclusive) is $1$.

The sets contain 49.999r% of the continuum respectively...Please do not answer '0.999r = 1'

As discussed in the comments, using "r" to signify a repeating decimal is very rare and a little confusing, but your intent with that is relatively clear. But your usage of repeating decimals and the word "continuum" make your question very hard to understand.

Here in math, as in philosophy, you should make it clear what your definitions of your terms are. By default, $49.999\ldots\%=50\%$. So if you meant something specific other than the standard meaning of decimal notation, you should make that clear if you want to be understood.

Similarly, "continuum" has a few closely related standard meanings in mathematics (1, 2, and more rarely: 3), but if you're working in a number system with infinitesimals, I think none of those three meanings could possibly apply. I think you were using "the continuum" to mean "the interval $[0,1]$ in an unspecified number system", but I am not sure.

Perhaps most importantly of all for this part of your question, how are you measuring "49.999r% of the...". It seems you something similar to length, but for which the result for $[0,1)$ is less than that for $[0,1]$.

Towards an Answer

If your only purpose is comparing sets like intervals, maybe comparing by subset inclusion suffices for your needs? If $\preceq$ and $\prec$ are used suggestively for subsets and proper subsets instead of $\subseteq$ and $\subsetneqq$, then certainly $[0,1/2)\cup[1/2,1)=[0,1)\prec[0,1]$.

But because you are writing things like "49.999r%", I think you want to distinguish the intervals $[0,1)$ and $[0,1]$ with numbers in a number system. Since such a thing would conflict with the standard definition(s) of length, I'll call it "pseudolength".

I imagine you want it to be translation invariant, and to assign $0$ to the empty set and positive quantities to other sets. You certainly need more numbers than reals if you'd like to assign a pseudolength of $x$ to $[0,x]$ (or $[0,x)$, whichever) for each real number $x$, since you don't want to give both the closed and half-open interval the same pseudolength.

It also seems you want the pseudolength of a finite union of disjoint sets to be the sum of the pseudolengths of the pieces (so pseudolengths are not just ordered, but can be added).

Taking this all together, if the psuedolength of $[1,1]$ is $\varepsilon$, and the pseudolength of $(0,1)$ is $x$, then the pseudolength of $[0,1)$ is $x+\varepsilon$ and that of $[0,1]$ is $x+2\varepsilon$. And if the pseudolength of $(0,1/2)$ is $y$, then we have the pseudolength of $[0,1/2)$ is $y+\varepsilon$. And so we are forced to conclude $2(y+\varepsilon)=x+\varepsilon$ so $y=x/2-\varepsilon/2$, apparently. This suggests the pseudolength of $[0,1/2)$ is $x/2+\varepsilon/2$, which may or may not be satisfying to you.