If this was true it would imply Set Theory embraces Paradox at its foundation so what am I not seeing?

If you have any countable infinite set you can relate any possible subset (and all possible subsets make up its Power Set) to a binary representation of the base two continuum [0,1] where the the digit to the right of the “decimal” point for a base two “decimal” is related to the order the countable infinite sets elements are shown with a 1 in that “decimal” place of that element if it is in the subset and a zero if that element is not in that subset. That is messy so I will show an example.

If you have the countable infinite set {1,2,3,4,5,6, ... } you can relate the finite set {1,3,5} with the base two “decimal” 0.10101 because you have 1’s in the first, third and fifth “decimal” place with zeros assumed in all the other infinite “decimal” places. So if a countable infinite set has any element in any given subset the “decimal” digit in a base two point in the base two “decimal continuum” would have a 1 corresponding to the place the element has in the original infinite set.

Note that also works for any countable set like {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z} and the whole set would be equal to the binary {0.11111111111111111111111111} with 26 ones for a power set that would contain 2^26 - 1 non empty subsets and 1 empty subset for 2^26 possible subsets. A specific subset of that sets Power Set could be {A,C,E} that would again go to a binary 0.10101 because you have 1’s in the first, third and fifth “decimal” place with zeros assumed in all the other 21 “decimal” places allowed so with 26 places you would have 2^26 possible subsets that are represented by all the base two 26 digits with zero being equivalent to the empty set but you would actually go from 0 to (2^26-1)/(2^26) for those 26 ones needed in the “decimal” places. You can order every possible subset and place all the possible subsets (again also known as its power set) in a bijection (not a surjection) with the finite counting numbers for any FINITE set.

So for the countable infinite set {1,2,3,4,5,6, ... } you can say that the value of 2/3 converted to binary or 0.101010... where those 10's repeated forever represents the set of all odd counting numbers and 1/3 as 0.010101... represents the evens so you can say the sum of the odd counting numbers and the even counting numbers gives the counting numbers or 2/3 + 1/3 = 1 where that “1” represents the whole set that is also represented by 0.101010... + 0.010101... = 0.111111... to see that as sums of infinite sets.

Question Why can you assume the “decimal” equivalency of saying that in base two 0.1 = 0.011111... with 1’s repeating forever would be true for the Power Set surjection to the base two “decimal continuum” [0,1] because that would demand that {1} = {2,3,4,5,6,7, ... } or that a finite set was equivalent to an infinite set ...

To me those are two very distinguishable sets yet it seems that “decimal equivalence” seems to demand they are the same. So does Set Theory demand a finite set is the same as an infinite set?

So can you use that decimal equivalency of demanding 0.0111111... = 0.1 in base two with Power Sets or in Cantor’s Diagonal Slash? Would that not demand that {1} = {2,3,4,5,6,7, ... }?

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    $\begingroup$ This map is surjection, but it is not a bijection, because it is not injective. In particular, as you have noticed, the sets $\{2,3,4,5,\ldots\}$ and $\{1\}$ "map" to the same number $0.1_2=0.5_{10}$. $\endgroup$ – Stinking Bishop Jan 30 at 19:39
  • $\begingroup$ But would not saying that {1} is not the same as {2,3,4,5,6,7, ... } mean that 0.1 is not equal to 0.0111111... so 0.0111111... could be an irrational number NEXT to 0.1? 0.0111111... does require an infinite number of digits to describe it... $\endgroup$ – D. Allred Jan 30 at 20:00
  • $\begingroup$ @D.Allred No, not at all - you're implicitly assuming that your representation function is injective, which it's not. All you have is a surjection $F:\mathcal{P}(\mathbb{N})\rightarrow [0,1]$; in general, $F(x)=F(y)$ does not imply that $x=y$ (and this is one such example). $\endgroup$ – Noah Schweber Jan 30 at 20:02
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    $\begingroup$ No. $\{2,3,4,5,\ldots\}$ is distinct from $\{1\}$ but $0.011111\ldots_2$ is the same as $0.1_2$. You seem to implicitly assume that this surjection is a bijection, i.e. that different objects must map to different objects. This is just not true here. I have noticed you are using the word "equivalency" [sic] but there is no equivalence here - the map from $\mathcal{P}(\mathbb N)$ to $[0,1]$ does not have an inverse in this case. $\endgroup$ – Stinking Bishop Jan 30 at 20:02

Your concern boils down to the following:

We have a map $F:\mathcal{P}(\mathbb{N})\rightarrow [0,1]$ with the property that $F(\{1\})=F(\{2,3,4,...\})$. Why doesn't this imply $\{1\}=\{2,3,4,...\}$?

Well, the answer is that there's no claim anywhere that the map $F$ is an injection. There are indeed bijections between $\mathcal{P}(\mathbb{N})$ and $[0,1]$, but they're more complicated than the $m$ you're looking at.

In fact, what you're observing is simply that $F$ is not injective. And this doesn't contradict anything - why should $F$ be injective?

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  • $\begingroup$ I am over my head in this so NOW I get to study "injection". Thanks for the Info. I take it that m( ) means measure so I will have to look at Measure Theory too. My time at the Library Computer is up so I will look for any other responses tomorrow. Thanks again. $\endgroup$ – D. Allred Jan 30 at 20:10
  • $\begingroup$ @D.Allred No, $m$ is not measure - it's just the representation map you're talking about. I've changed the name to avoid confusion. $\endgroup$ – Noah Schweber Jan 30 at 20:11

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