Formulation of recursive subset equality

  • Thread starter Thread starter thegluups
  • Start date Start date
thegluups
Messages
1
Reaction score
0
Hi,

I'm currently trying to express the following:

given a structured set of sets, of the form {1}, {1,{1,1}}, {{1,1},{1,1}}, or {1,{{1,1},1}} etc.

I want to be able to test whether two sets are reductions of each other, where reduction means, that:
{1,{1,1} is equal to {1}, or {1,1}, but not {{1,1},1}.

The logic behind reduction is that I can reduce a subset to a subset of smaller cardinality, or singleton set (so {1,1} is a reduction of {1,{1,1}}. The relation is transitive, so given {1} is a reduction of {1,1} then {1} is a reduction of {1,{1,1}}.

I think I have the right properties: given two ordered sets A and B, B is a reduction of A

1) iff A and B contain the same number of subsets, and there is a bijection between the subsets of A and the subsets of B, where the corresponding subset in B is itself a reduction of the corresponding subset in A. (basically what I'm trying to say is that if you have {1,2,3} and {a,b,c} then 1 is linked to a, 2 to b, etc. also works for subsets, where you have {{1,2},3,4} and {a,b,c} where in that case {1,2} is linked with a, 3 with b, and 4 with c)
2) iff there exists a set R such that B is a reduction of R, and R is a reduction of A.

Does this definition make sense? And if yes, I'm really struggling to provide formal notation for 1. I'm not very familiar with set notation.

Thanks!
 
Physics news on Phys.org
thegluups said:
Hi,

I'm currently trying to express the following:

given a structured set of sets, of the form {1}, {1,{1,1}}, {{1,1},{1,1}}, or {1,{{1,1},1}} etc.

I want to be able to test whether two sets are reductions of each other, where reduction means, that:
{1,{1,1} is equal to {1}, or {1,1}, but not {{1,1},1}.

The logic behind reduction is that I can reduce a subset to a subset of smaller cardinality, or singleton set (so {1,1} is a reduction of {1,{1,1}}. The relation is transitive, so given {1} is a reduction of {1,1} then {1} is a reduction of {1,{1,1}}.

I think I have the right properties: given two ordered sets A and B, B is a reduction of A

1) iff A and B contain the same number of subsets, and there is a bijection between the subsets of A and the subsets of B, where the corresponding subset in B is itself a reduction of the corresponding subset in A. (basically what I'm trying to say is that if you have {1,2,3} and {a,b,c} then 1 is linked to a, 2 to b, etc. also works for subsets, where you have {{1,2},3,4} and {a,b,c} where in that case {1,2} is linked with a, 3 with b, and 4 with c)
2) iff there exists a set R such that B is a reduction of R, and R is a reduction of A.

Does this definition make sense? And if yes, I'm really struggling to provide formal notation for 1. I'm not very familiar with set notation.

Thanks!

The definition of a set is a collection with no repetitions and in which order doesn't matter. What you are dealing with are called "sequences." These "reductions" are subsequences of sequences.

You may define words however you like, but nonstandard definitions are a pointless headache for the reader.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...

Similar threads

Replies
1
Views
1K
Replies
5
Views
1K
Replies
57
Views
7K
Replies
6
Views
2K
Replies
4
Views
3K
Replies
2
Views
2K
Back
Top