Does $pi$ contain every real?












2












$begingroup$


One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.



$pi = 3.141592653589793238462643383279502884197169399...$



for instance, for $m=3$, $n=5$, this will be



3.141592653589793238462643383279502884197169399...



giving



$0.469363019...$



I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.



So I guess this method can't create any real. But why?






elementary-set-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
    $endgroup$
    – mathworker21
    Dec 12 '18 at 8:57






  • 1




    $begingroup$
    The number of pairs $n, : k$ is countable!
    $endgroup$
    – R.C.Cowsik
    Dec 12 '18 at 8:58








  • 4




    $begingroup$
    I think the more interesting question is : give an example of a real number that cannot be produced like this.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 9:07










  • $begingroup$
    My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:36










  • $begingroup$
    @SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
    $endgroup$
    – Holo
    Dec 12 '18 at 9:53
















2












$begingroup$


One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.



$pi = 3.141592653589793238462643383279502884197169399...$



for instance, for $m=3$, $n=5$, this will be



3.141592653589793238462643383279502884197169399...



giving



$0.469363019...$



I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.



So I guess this method can't create any real. But why?






elementary-set-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
    $endgroup$
    – mathworker21
    Dec 12 '18 at 8:57






  • 1




    $begingroup$
    The number of pairs $n, : k$ is countable!
    $endgroup$
    – R.C.Cowsik
    Dec 12 '18 at 8:58








  • 4




    $begingroup$
    I think the more interesting question is : give an example of a real number that cannot be produced like this.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 9:07










  • $begingroup$
    My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:36










  • $begingroup$
    @SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
    $endgroup$
    – Holo
    Dec 12 '18 at 9:53














2












2








2





$begingroup$


One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.



$pi = 3.141592653589793238462643383279502884197169399...$



for instance, for $m=3$, $n=5$, this will be



3.141592653589793238462643383279502884197169399...



giving



$0.469363019...$



I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.



So I guess this method can't create any real. But why?






elementary-set-theory







share|cite|improve this question











$endgroup$




One can generate decimal expansions between $0$ and $1$ by taking with the $m$th digit of $pi$, and then taking every $n$th digit.



$pi = 3.141592653589793238462643383279502884197169399...$



for instance, for $m=3$, $n=5$, this will be



3.141592653589793238462643383279502884197169399...



giving



$0.469363019...$



I can't see why this shouldn't be able to create every real number. But if this creates any real number given a pair if natural numbers $m$ and $n$, it would mean that the cardinality of the continuum is equal to that of the natural numbers.



So I guess this method can't create any real. But why?






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 18:14









verret

3,3161923




3,3161923










asked Dec 12 '18 at 8:52









MrMartinMrMartin

757




757








  • 3




    $begingroup$
    You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
    $endgroup$
    – mathworker21
    Dec 12 '18 at 8:57






  • 1




    $begingroup$
    The number of pairs $n, : k$ is countable!
    $endgroup$
    – R.C.Cowsik
    Dec 12 '18 at 8:58








  • 4




    $begingroup$
    I think the more interesting question is : give an example of a real number that cannot be produced like this.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 9:07










  • $begingroup$
    My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:36










  • $begingroup$
    @SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
    $endgroup$
    – Holo
    Dec 12 '18 at 9:53














  • 3




    $begingroup$
    You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
    $endgroup$
    – mathworker21
    Dec 12 '18 at 8:57






  • 1




    $begingroup$
    The number of pairs $n, : k$ is countable!
    $endgroup$
    – R.C.Cowsik
    Dec 12 '18 at 8:58








  • 4




    $begingroup$
    I think the more interesting question is : give an example of a real number that cannot be produced like this.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 12 '18 at 9:07










  • $begingroup$
    My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
    $endgroup$
    – SmileyCraft
    Dec 12 '18 at 9:36










  • $begingroup$
    @SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
    $endgroup$
    – Holo
    Dec 12 '18 at 9:53








3




3




$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57




$begingroup$
You literally answered your own question. Also, it's unknown whether the decimal expansion of $pi$ becomes only 1's and 0's after a certain point, so you definitely can't prove as of now that every real number occurs as a subsequence of the digits of $pi$, let alone in the manner you described.
$endgroup$
– mathworker21
Dec 12 '18 at 8:57




1




1




$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58






$begingroup$
The number of pairs $n, : k$ is countable!
$endgroup$
– R.C.Cowsik
Dec 12 '18 at 8:58






4




4




$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07




$begingroup$
I think the more interesting question is : give an example of a real number that cannot be produced like this.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 12 '18 at 9:07












$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36




$begingroup$
My guess is that this process never creates a rational, as that would require infinite repetition. I don't know if this can be proven using the current tools though.
$endgroup$
– SmileyCraft
Dec 12 '18 at 9:36












$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53




$begingroup$
@SmileyCraft I think it is possible to prove that infinite number of rational numbers cannot be created(although I didn't done it), but proving that no rational is there sounds hard
$endgroup$
– Holo
Dec 12 '18 at 9:53










1 Answer
1






active

oldest

votes


















5












$begingroup$

You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.



In particular, the argument would go like so:




  1. Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.

  2. Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

  3. Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$

  4. Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$

  5. Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

  6. From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.

  7. Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:20










  • $begingroup$
    @MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:24










  • $begingroup$
    @MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 9:30










  • $begingroup$
    Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:32






  • 3




    $begingroup$
    @MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:40












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.



In particular, the argument would go like so:




  1. Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.

  2. Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

  3. Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$

  4. Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$

  5. Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

  6. From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.

  7. Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:20










  • $begingroup$
    @MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:24










  • $begingroup$
    @MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 9:30










  • $begingroup$
    Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:32






  • 3




    $begingroup$
    @MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:40
















5












$begingroup$

You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.



In particular, the argument would go like so:




  1. Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.

  2. Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

  3. Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$

  4. Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$

  5. Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

  6. From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.

  7. Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:20










  • $begingroup$
    @MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:24










  • $begingroup$
    @MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 9:30










  • $begingroup$
    Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:32






  • 3




    $begingroup$
    @MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:40














5












5








5





$begingroup$

You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.



In particular, the argument would go like so:




  1. Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.

  2. Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

  3. Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$

  4. Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$

  5. Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

  6. From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.

  7. Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.






share|cite|improve this answer











$endgroup$



You basically present a mapping from $mathbb N^2$ to $mathbb R$ and ask if the mapping is surjective. Since $mathbb N^2$ is countable, and $mathbb R$ is uncountable, a simple diagonalization argument can be made to show that the mapping cannot be surjective.



In particular, the argument would go like so:




  1. Let $F$ be the mapping you construct going from $mathbb N^2$ to $[0,1]$. That is, $F(m,n)$ is the number obtained by taking with the $m$th digit of $pi$, and then taking every $n$th digit.

  2. Let $f:mathbb Ntomathbb N^2$ be a bijective function. (we know such a function exists, and it doesn't really matter which one we take) and let $f_1,f_2$ be its components, i.e. $f(n)=(f_1(n), f_2(n))$.

  3. Denote the digits of $F(f(n))$ as $a_i^{(n)}$, that is, $F(f(n)) = 0.a_1^{(n)}a_2^{(n)}a_3^{(n)}dots$

  4. Let $$b_i=begin{cases}1 & text{if } a_i^{(i)} = 0\ 0& text{if } a_i^{(i)} neq 0end{cases}$$

  5. Let $b=0.b_1b_2b_3dots$. Clearly, for all $ninmathbb N$, we see that $F(f(n))neq b$ because the $n$-th digit of $F(f(n))$ is not the same as the $n$-th digit of $b$.

  6. From $5$, it follows that the number $b$ is not in the domain of $Fcirc f$.

  7. Because $f$ is bijective, we conclude that $b$ is not in the domain of $F$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 9:37

























answered Dec 12 '18 at 9:05









5xum5xum

91.8k394161




91.8k394161












  • $begingroup$
    Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:20










  • $begingroup$
    @MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:24










  • $begingroup$
    @MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 9:30










  • $begingroup$
    Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:32






  • 3




    $begingroup$
    @MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:40


















  • $begingroup$
    Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:20










  • $begingroup$
    @MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:24










  • $begingroup$
    @MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
    $endgroup$
    – Yves Daoust
    Dec 12 '18 at 9:30










  • $begingroup$
    Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
    $endgroup$
    – MrMartin
    Dec 12 '18 at 9:32






  • 3




    $begingroup$
    @MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
    $endgroup$
    – 5xum
    Dec 12 '18 at 9:40
















$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20




$begingroup$
Thank you for the clean re-wording. The diagonalization argument will fail when you can show that the number is just further down the list, as in this case
$endgroup$
– MrMartin
Dec 12 '18 at 9:20












$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24




$begingroup$
@MrMartin The diagonalization argument can show that any mapping from a countable set to $mathbb R$ cannot be surjective. It will not fail. It is possible to construct a number that is not in your mapping.
$endgroup$
– 5xum
Dec 12 '18 at 9:24












$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30




$begingroup$
@MrMartin: how can you draw any conclusion about a diagonalization argument without even seeing it ? Or you need to prove that no such argument can be written.
$endgroup$
– Yves Daoust
Dec 12 '18 at 9:30












$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32




$begingroup$
Following this line of reasoning, the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself.
$endgroup$
– MrMartin
Dec 12 '18 at 9:32




3




3




$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40




$begingroup$
@MrMartin Happy to help. Remember to accept the answer if it is what you needed. And a piece of general advice: avoid giving sweeping statements like "the diagonalization argument could also be applied to show that the cardinality of $mathbb R$ is greater than itself. " before trying to prove them. It makes you sound like you get an answer and immediatelly assume it is wrong. The aggressive tone may not go down well with everyone.
$endgroup$
– 5xum
Dec 12 '18 at 9:40


















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