On the construction of real numbers…

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Can somebody give me some intuitive reasoning as to why there should be (and is) more equivalences classes of pairwise Cauchy sequences of rational numbers than there are rational numbers?

Also, from this perspective is there a way to see that the cardinality of this collection of equivalence classes is $2^aleph$? I understand that this is true from then bijecting these equivalence classes with the real numbers, then a bijection with $(0,1)$ and then expressing each decimal in binary, but I was wondering if I could get another perspective.

real-analysis real-numbers cauchy-sequences

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edited Jul 22 at 21:27

José Carlos Santos

113k1698177

asked Jul 22 at 13:43

Michael Vaughan

29711

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
  – Thomas
  Jul 22 at 14:03
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

2

Can somebody give me some intuitive reasoning as to why there should be (and is) more equivalences classes of pairwise Cauchy sequences of rational numbers than there are rational numbers?

Also, from this perspective is there a way to see that the cardinality of this collection of equivalence classes is $2^aleph$? I understand that this is true from then bijecting these equivalence classes with the real numbers, then a bijection with $(0,1)$ and then expressing each decimal in binary, but I was wondering if I could get another perspective.

real-analysis real-numbers cauchy-sequences

share|cite|improve this question

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

asked Jul 22 at 13:43

Michael Vaughan

29711

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
  – Thomas
  Jul 22 at 14:03
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

2

up vote
1
down vote

favorite

2

2

Can somebody give me some intuitive reasoning as to why there should be (and is) more equivalences classes of pairwise Cauchy sequences of rational numbers than there are rational numbers?

Also, from this perspective is there a way to see that the cardinality of this collection of equivalence classes is $2^aleph$? I understand that this is true from then bijecting these equivalence classes with the real numbers, then a bijection with $(0,1)$ and then expressing each decimal in binary, but I was wondering if I could get another perspective.

real-analysis real-numbers cauchy-sequences

share|cite|improve this question

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

asked Jul 22 at 13:43

Michael Vaughan

29711

Can somebody give me some intuitive reasoning as to why there should be (and is) more equivalences classes of pairwise Cauchy sequences of rational numbers than there are rational numbers?

Also, from this perspective is there a way to see that the cardinality of this collection of equivalence classes is $2^aleph$? I understand that this is true from then bijecting these equivalence classes with the real numbers, then a bijection with $(0,1)$ and then expressing each decimal in binary, but I was wondering if I could get another perspective.

real-analysis real-numbers cauchy-sequences

share|cite|improve this question

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

asked Jul 22 at 13:43

Michael Vaughan

29711

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

edited Jul 22 at 21:27

José Carlos Santos

113k1698177

113k1698177

asked Jul 22 at 13:43

Michael Vaughan

29711

asked Jul 22 at 13:43

Michael Vaughan

29711

asked Jul 22 at 13:43

Michael Vaughan

29711

29711

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
  – Thomas
  Jul 22 at 14:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
  – Thomas
  Jul 22 at 14:03
  

  
  
  
  

1

1

do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
– Thomas
Jul 22 at 14:03

do you know Cantor's diagonal argument? en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
– Thomas
Jul 22 at 14:03

add a comment | 

2 Answers
2

active

oldest

votes

up vote
3
down vote



accepted

For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence
$(b_n)$ by $b_n=sum_k=1^n a_k/3^k$. Then each $(b_n)$ is a Cauchy
sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are
uncountably many $(a_n)$.

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why did you need to have a $3^k$ in the denominator rather than $2^k$?
  – Michael Vaughan
  Jul 22 at 19:45
  

  
  
  
  

add a comment | 

up vote
2
down vote

Let $mathcal Q$ be the set of subsets $A$ of $mathbb Q$ such that:

1. $Aneqemptyset$;


2. if $ain A$, then there's a $bin A$ such that $a<b$;


3. $A$ has an upper bound.

For each $Ainmathcal Q$, consider a Cauchy sequence $(q_n)_ninmathbb N$ of elements of $mathbb Q$ such that, for each $ninmathbb N$,

1. there is some $ain A$ such that $ageqslant q_n$;


2. $q_n+frac1n$ is an upper bound of $A$.

Then, if $Q,Q'inmathcal Q$ are such that there is a $qin Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $mathcal Q$, you consider the equivalence relation$$Qsim Q'iff (forall qin Q)(exists q'in Q'):qgeqslant q'text and vice-versa,$$then there are (at least) as many equivalence classes with respect to $sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $mathcal Q$ with respect to $sim$ is, at least, $2^aleph_0$.

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence
$(b_n)$ by $b_n=sum_k=1^n a_k/3^k$. Then each $(b_n)$ is a Cauchy
sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are
uncountably many $(a_n)$.

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why did you need to have a $3^k$ in the denominator rather than $2^k$?
  – Michael Vaughan
  Jul 22 at 19:45
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence
$(b_n)$ by $b_n=sum_k=1^n a_k/3^k$. Then each $(b_n)$ is a Cauchy
sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are
uncountably many $(a_n)$.

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why did you need to have a $3^k$ in the denominator rather than $2^k$?
  – Michael Vaughan
  Jul 22 at 19:45
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence
$(b_n)$ by $b_n=sum_k=1^n a_k/3^k$. Then each $(b_n)$ is a Cauchy
sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are
uncountably many $(a_n)$.

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

For each sequence $(a_n)$ of zeros and ones, define the Cauchy sequence
$(b_n)$ by $b_n=sum_k=1^n a_k/3^k$. Then each $(b_n)$ is a Cauchy
sequence and different $(a_n)$ yield inequivalent $(b_n)$. There are
uncountably many $(a_n)$.

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 14:12

Lord Shark the Unknown

85.2k950111

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why did you need to have a $3^k$ in the denominator rather than $2^k$?
  – Michael Vaughan
  Jul 22 at 19:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why did you need to have a $3^k$ in the denominator rather than $2^k$?
  – Michael Vaughan
  Jul 22 at 19:45
  

  
  
  
  

Why did you need to have a $3^k$ in the denominator rather than $2^k$?
– Michael Vaughan
Jul 22 at 19:45

Why did you need to have a $3^k$ in the denominator rather than $2^k$?
– Michael Vaughan
Jul 22 at 19:45

add a comment | 

up vote
2
down vote

Let $mathcal Q$ be the set of subsets $A$ of $mathbb Q$ such that:

1. $Aneqemptyset$;


2. if $ain A$, then there's a $bin A$ such that $a<b$;


3. $A$ has an upper bound.

For each $Ainmathcal Q$, consider a Cauchy sequence $(q_n)_ninmathbb N$ of elements of $mathbb Q$ such that, for each $ninmathbb N$,

1. there is some $ain A$ such that $ageqslant q_n$;


2. $q_n+frac1n$ is an upper bound of $A$.

Then, if $Q,Q'inmathcal Q$ are such that there is a $qin Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $mathcal Q$, you consider the equivalence relation$$Qsim Q'iff (forall qin Q)(exists q'in Q'):qgeqslant q'text and vice-versa,$$then there are (at least) as many equivalence classes with respect to $sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $mathcal Q$ with respect to $sim$ is, at least, $2^aleph_0$.

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

add a comment | 

up vote
2
down vote

Let $mathcal Q$ be the set of subsets $A$ of $mathbb Q$ such that:

1. $Aneqemptyset$;


2. if $ain A$, then there's a $bin A$ such that $a<b$;


3. $A$ has an upper bound.

For each $Ainmathcal Q$, consider a Cauchy sequence $(q_n)_ninmathbb N$ of elements of $mathbb Q$ such that, for each $ninmathbb N$,

1. there is some $ain A$ such that $ageqslant q_n$;


2. $q_n+frac1n$ is an upper bound of $A$.

Then, if $Q,Q'inmathcal Q$ are such that there is a $qin Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $mathcal Q$, you consider the equivalence relation$$Qsim Q'iff (forall qin Q)(exists q'in Q'):qgeqslant q'text and vice-versa,$$then there are (at least) as many equivalence classes with respect to $sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $mathcal Q$ with respect to $sim$ is, at least, $2^aleph_0$.

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

add a comment | 

up vote
2
down vote

up vote
2
down vote

Let $mathcal Q$ be the set of subsets $A$ of $mathbb Q$ such that:

1. $Aneqemptyset$;


2. if $ain A$, then there's a $bin A$ such that $a<b$;


3. $A$ has an upper bound.

For each $Ainmathcal Q$, consider a Cauchy sequence $(q_n)_ninmathbb N$ of elements of $mathbb Q$ such that, for each $ninmathbb N$,

1. there is some $ain A$ such that $ageqslant q_n$;


2. $q_n+frac1n$ is an upper bound of $A$.

Then, if $Q,Q'inmathcal Q$ are such that there is a $qin Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $mathcal Q$, you consider the equivalence relation$$Qsim Q'iff (forall qin Q)(exists q'in Q'):qgeqslant q'text and vice-versa,$$then there are (at least) as many equivalence classes with respect to $sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $mathcal Q$ with respect to $sim$ is, at least, $2^aleph_0$.

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

Let $mathcal Q$ be the set of subsets $A$ of $mathbb Q$ such that:

1. $Aneqemptyset$;


2. if $ain A$, then there's a $bin A$ such that $a<b$;


3. $A$ has an upper bound.

For each $Ainmathcal Q$, consider a Cauchy sequence $(q_n)_ninmathbb N$ of elements of $mathbb Q$ such that, for each $ninmathbb N$,

1. there is some $ain A$ such that $ageqslant q_n$;


2. $q_n+frac1n$ is an upper bound of $A$.

Then, if $Q,Q'inmathcal Q$ are such that there is a $qin Q$ which is an upper bound of $Q'$ (or vice-versa), then a Cauchy sequence obtained from $Q$ and a Cauchy sequence obtained from $Q'$ cannot belong to the same equivalence class.

So if, in $mathcal Q$, you consider the equivalence relation$$Qsim Q'iff (forall qin Q)(exists q'in Q'):qgeqslant q'text and vice-versa,$$then there are (at least) as many equivalence classes with respect to $sim$ as there are classes of equivalence of Cauchy sequences. But it's not hard to prove that the number of equivalence classes of $mathcal Q$ with respect to $sim$ is, at least, $2^aleph_0$.

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

answered Jul 22 at 14:13

José Carlos Santos

113k1698177

113k1698177

add a comment | 

add a comment | 

 

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