Creating an infinite list of sets of natural numbers.

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
-1
down vote

favorite












We can define a list like this, we begin with two elements in rows -1 and 1



(-1) $colorredvarnothing$



(1) $colorblue1$



Then we copy the top half of our list and paste it on the bottom and vice versa



(-2) $colorblue1$



(-1) $colorredvarnothing$



(1) $colorblue1$



(2) $colorredvarnothing$



Then we append the top with empty sets and the bottom with the next natural number



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



Again we copy the top half of our list and paste it on the bottom and vice versa



(-4) $colorblue1,2$



(-3) $colorbluevarnothing,2$



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



(3) $colorred1,varnothing$



(4) $colorredvarnothing,varnothing$



Again we append the top with empty sets and the bottom with the next natural number



(-4) $colorred1,2,varnothing$



(-3) $colorredvarnothing,2,varnothing$



(-2) $colorred1,varnothing,varnothing$



(-1) $colorredvarnothing,varnothing,varnothing$



(1) $colorblue1,2,3$



(2) $colorbluevarnothing,2,3$



(3) $colorblue1,varnothing,3$



(4) $colorbluevarnothing,varnothing,3$



If we do this forever, will the set of natural numbers be on some row in this list? I would think "yes, it is in row (1)".



It is obvious that there is a symmetry between the top half of the list and the bottom half of the list, but will this symmetry hold when there is an infinite number of elements in each row? Will the cardinality of the top half of the list be the same as the cardinality of the bottom half of the list?



I will add a note to explain why I find this question interesting.



After three steps our list has 8 rows where four (-1) to (-4) elements are in the top half and four (1) to (4) elements are in the bottom half, so the cardinality of the top is the same as the bottom. And we have all the subsets of the set 1,2,3 represented on our list.



If we take one more step, we will have 16 rows, where 8 are in top half of our list and 8 in the bottom. And for any number of steps we will have the same number of rows in the top as we have in the bottom and all the possible subsets will be represented. This is just the natural outcome from the way we construct our list.



If the cardinality of the top half of the list is still the same as the bottom of the list when we have an infinite number of rows in our list, we will have a problem.



All the sets in the top half of our list will have a finite number of elements and all the sets in the bottom of the list will have an infinite number of elements. This would contradict Cantor's idea of infinite sets, if all the subsets of the natural numbers are present on this list.



If we could find an infinite set that is not in the bottom half of our list, we could preserve Cantor. But we can't find it by diagonalizing the bottom half of the list. (diagonalization will only produce finite sets)



So, how do we harmonize this list with Cantor's diagonal argument?




elementary-set-theory




share|cite|improve this question





















  • the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
    – Rushabh Mehta
    Aug 6 at 15:18






  • 1




    Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
    – Asaf Karagila♦
    Aug 6 at 16:11











  • @Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
    – Ivan Hieno
    Aug 7 at 12:50










  • @Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
    – Ivan Hieno
    Aug 8 at 12:38










  • I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
    – pregunton
    Aug 8 at 12:43














up vote
-1
down vote

favorite












We can define a list like this, we begin with two elements in rows -1 and 1



(-1) $colorredvarnothing$



(1) $colorblue1$



Then we copy the top half of our list and paste it on the bottom and vice versa



(-2) $colorblue1$



(-1) $colorredvarnothing$



(1) $colorblue1$



(2) $colorredvarnothing$



Then we append the top with empty sets and the bottom with the next natural number



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



Again we copy the top half of our list and paste it on the bottom and vice versa



(-4) $colorblue1,2$



(-3) $colorbluevarnothing,2$



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



(3) $colorred1,varnothing$



(4) $colorredvarnothing,varnothing$



Again we append the top with empty sets and the bottom with the next natural number



(-4) $colorred1,2,varnothing$



(-3) $colorredvarnothing,2,varnothing$



(-2) $colorred1,varnothing,varnothing$



(-1) $colorredvarnothing,varnothing,varnothing$



(1) $colorblue1,2,3$



(2) $colorbluevarnothing,2,3$



(3) $colorblue1,varnothing,3$



(4) $colorbluevarnothing,varnothing,3$



If we do this forever, will the set of natural numbers be on some row in this list? I would think "yes, it is in row (1)".



It is obvious that there is a symmetry between the top half of the list and the bottom half of the list, but will this symmetry hold when there is an infinite number of elements in each row? Will the cardinality of the top half of the list be the same as the cardinality of the bottom half of the list?



I will add a note to explain why I find this question interesting.



After three steps our list has 8 rows where four (-1) to (-4) elements are in the top half and four (1) to (4) elements are in the bottom half, so the cardinality of the top is the same as the bottom. And we have all the subsets of the set 1,2,3 represented on our list.



If we take one more step, we will have 16 rows, where 8 are in top half of our list and 8 in the bottom. And for any number of steps we will have the same number of rows in the top as we have in the bottom and all the possible subsets will be represented. This is just the natural outcome from the way we construct our list.



If the cardinality of the top half of the list is still the same as the bottom of the list when we have an infinite number of rows in our list, we will have a problem.



All the sets in the top half of our list will have a finite number of elements and all the sets in the bottom of the list will have an infinite number of elements. This would contradict Cantor's idea of infinite sets, if all the subsets of the natural numbers are present on this list.



If we could find an infinite set that is not in the bottom half of our list, we could preserve Cantor. But we can't find it by diagonalizing the bottom half of the list. (diagonalization will only produce finite sets)



So, how do we harmonize this list with Cantor's diagonal argument?




elementary-set-theory




share|cite|improve this question





















  • the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
    – Rushabh Mehta
    Aug 6 at 15:18






  • 1




    Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
    – Asaf Karagila♦
    Aug 6 at 16:11











  • @Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
    – Ivan Hieno
    Aug 7 at 12:50










  • @Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
    – Ivan Hieno
    Aug 8 at 12:38










  • I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
    – pregunton
    Aug 8 at 12:43












up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











We can define a list like this, we begin with two elements in rows -1 and 1



(-1) $colorredvarnothing$



(1) $colorblue1$



Then we copy the top half of our list and paste it on the bottom and vice versa



(-2) $colorblue1$



(-1) $colorredvarnothing$



(1) $colorblue1$



(2) $colorredvarnothing$



Then we append the top with empty sets and the bottom with the next natural number



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



Again we copy the top half of our list and paste it on the bottom and vice versa



(-4) $colorblue1,2$



(-3) $colorbluevarnothing,2$



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



(3) $colorred1,varnothing$



(4) $colorredvarnothing,varnothing$



Again we append the top with empty sets and the bottom with the next natural number



(-4) $colorred1,2,varnothing$



(-3) $colorredvarnothing,2,varnothing$



(-2) $colorred1,varnothing,varnothing$



(-1) $colorredvarnothing,varnothing,varnothing$



(1) $colorblue1,2,3$



(2) $colorbluevarnothing,2,3$



(3) $colorblue1,varnothing,3$



(4) $colorbluevarnothing,varnothing,3$



If we do this forever, will the set of natural numbers be on some row in this list? I would think "yes, it is in row (1)".



It is obvious that there is a symmetry between the top half of the list and the bottom half of the list, but will this symmetry hold when there is an infinite number of elements in each row? Will the cardinality of the top half of the list be the same as the cardinality of the bottom half of the list?



I will add a note to explain why I find this question interesting.



After three steps our list has 8 rows where four (-1) to (-4) elements are in the top half and four (1) to (4) elements are in the bottom half, so the cardinality of the top is the same as the bottom. And we have all the subsets of the set 1,2,3 represented on our list.



If we take one more step, we will have 16 rows, where 8 are in top half of our list and 8 in the bottom. And for any number of steps we will have the same number of rows in the top as we have in the bottom and all the possible subsets will be represented. This is just the natural outcome from the way we construct our list.



If the cardinality of the top half of the list is still the same as the bottom of the list when we have an infinite number of rows in our list, we will have a problem.



All the sets in the top half of our list will have a finite number of elements and all the sets in the bottom of the list will have an infinite number of elements. This would contradict Cantor's idea of infinite sets, if all the subsets of the natural numbers are present on this list.



If we could find an infinite set that is not in the bottom half of our list, we could preserve Cantor. But we can't find it by diagonalizing the bottom half of the list. (diagonalization will only produce finite sets)



So, how do we harmonize this list with Cantor's diagonal argument?




elementary-set-theory




share|cite|improve this question













We can define a list like this, we begin with two elements in rows -1 and 1



(-1) $colorredvarnothing$



(1) $colorblue1$



Then we copy the top half of our list and paste it on the bottom and vice versa



(-2) $colorblue1$



(-1) $colorredvarnothing$



(1) $colorblue1$



(2) $colorredvarnothing$



Then we append the top with empty sets and the bottom with the next natural number



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



Again we copy the top half of our list and paste it on the bottom and vice versa



(-4) $colorblue1,2$



(-3) $colorbluevarnothing,2$



(-2) $colorred1,varnothing$



(-1) $colorredvarnothing,varnothing$



(1) $colorblue1,2$



(2) $colorbluevarnothing,2$



(3) $colorred1,varnothing$



(4) $colorredvarnothing,varnothing$



Again we append the top with empty sets and the bottom with the next natural number



(-4) $colorred1,2,varnothing$



(-3) $colorredvarnothing,2,varnothing$



(-2) $colorred1,varnothing,varnothing$



(-1) $colorredvarnothing,varnothing,varnothing$



(1) $colorblue1,2,3$



(2) $colorbluevarnothing,2,3$



(3) $colorblue1,varnothing,3$



(4) $colorbluevarnothing,varnothing,3$



If we do this forever, will the set of natural numbers be on some row in this list? I would think "yes, it is in row (1)".



It is obvious that there is a symmetry between the top half of the list and the bottom half of the list, but will this symmetry hold when there is an infinite number of elements in each row? Will the cardinality of the top half of the list be the same as the cardinality of the bottom half of the list?



I will add a note to explain why I find this question interesting.



After three steps our list has 8 rows where four (-1) to (-4) elements are in the top half and four (1) to (4) elements are in the bottom half, so the cardinality of the top is the same as the bottom. And we have all the subsets of the set 1,2,3 represented on our list.



If we take one more step, we will have 16 rows, where 8 are in top half of our list and 8 in the bottom. And for any number of steps we will have the same number of rows in the top as we have in the bottom and all the possible subsets will be represented. This is just the natural outcome from the way we construct our list.



If the cardinality of the top half of the list is still the same as the bottom of the list when we have an infinite number of rows in our list, we will have a problem.



All the sets in the top half of our list will have a finite number of elements and all the sets in the bottom of the list will have an infinite number of elements. This would contradict Cantor's idea of infinite sets, if all the subsets of the natural numbers are present on this list.



If we could find an infinite set that is not in the bottom half of our list, we could preserve Cantor. But we can't find it by diagonalizing the bottom half of the list. (diagonalization will only produce finite sets)



So, how do we harmonize this list with Cantor's diagonal argument?





elementary-set-theory





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 8 at 12:23
























asked Aug 6 at 14:44









Ivan Hieno

1119




1119











  • the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
    – Rushabh Mehta
    Aug 6 at 15:18






  • 1




    Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
    – Asaf Karagila♦
    Aug 6 at 16:11











  • @Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
    – Ivan Hieno
    Aug 7 at 12:50










  • @Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
    – Ivan Hieno
    Aug 8 at 12:38










  • I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
    – pregunton
    Aug 8 at 12:43
















  • the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
    – Rushabh Mehta
    Aug 6 at 15:18






  • 1




    Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
    – Asaf Karagila♦
    Aug 6 at 16:11











  • @Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
    – Ivan Hieno
    Aug 7 at 12:50










  • @Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
    – Ivan Hieno
    Aug 8 at 12:38










  • I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
    – pregunton
    Aug 8 at 12:43















the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
– Rushabh Mehta
Aug 6 at 15:18




the Row 1 being natural numbers is quite obvious, while I don't understand what your other questions mean
– Rushabh Mehta
Aug 6 at 15:18




1




1




Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
– Asaf Karagila♦
Aug 6 at 16:11





Count. (1) $1$, (2) $2$, (3) $3$, ... (n) $n$, ... Did at some point you count all the natural numbers?
– Asaf Karagila♦
Aug 6 at 16:11













@Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
– Ivan Hieno
Aug 7 at 12:50




@Asaf Karagila Would the same apply to the set of even natural numbers, count (1) ∅, (2) 2, (3) ∅, (4) 4, (5) ∅, (6) 6, ... (n-1) ∅, (n) n, ... Did at some point you count all the even natural numbers?
– Ivan Hieno
Aug 7 at 12:50












@Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
– Ivan Hieno
Aug 8 at 12:38




@Rushabh Mehta I added a little information to the question to explain what I am trying to figure out here.
– Ivan Hieno
Aug 8 at 12:38












I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
– pregunton
Aug 8 at 12:43




I think there is no contradiction. When we pass to the limit, on the top half we have the finite subsets, as you say, and on the bottom half we have the cofinite subsets (subsets such that the complement is finite). An example of set that is not finite or cofinite, and thus is not in the final list, would be e.g. the set of even numbers.
– pregunton
Aug 8 at 12:43















active

oldest

votes











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873936%2fcreating-an-infinite-list-of-sets-of-natural-numbers%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873936%2fcreating-an-infinite-list-of-sets-of-natural-numbers%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
Read more