Why is the set of all real numbers uncountable?












4












$begingroup$


I understand Cantor's diagonal argument, but it just doesn't seem to jive for some reason.



Lets say I assign the following numbers ad infinitum...




  1. $1to 0.1$

  2. $2to 0.2$

  3. $3to 0.3$
    ...

  4. $10to 0.10$

  5. $11to 0.11$


and so on... How come there's supposedly at least one more real number than you can map to a member of $mathbb{N}$?










share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Because... Cantor explicitly tells us what that "one" real number is... ???
    $endgroup$
    – The Chaz 2.0
    Nov 14 '12 at 16:39






  • 6




    $begingroup$
    @agent154: By the way, where is $frac13$ in your list?
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:40








  • 2




    $begingroup$
    @agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:44








  • 6




    $begingroup$
    ...are you sure you understand Cantor's diagonal argument?
    $endgroup$
    – camccann
    Nov 14 '12 at 16:45






  • 5




    $begingroup$
    @MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
    $endgroup$
    – Thomas Andrews
    Nov 14 '12 at 16:45
















4












$begingroup$


I understand Cantor's diagonal argument, but it just doesn't seem to jive for some reason.



Lets say I assign the following numbers ad infinitum...




  1. $1to 0.1$

  2. $2to 0.2$

  3. $3to 0.3$
    ...

  4. $10to 0.10$

  5. $11to 0.11$


and so on... How come there's supposedly at least one more real number than you can map to a member of $mathbb{N}$?










share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Because... Cantor explicitly tells us what that "one" real number is... ???
    $endgroup$
    – The Chaz 2.0
    Nov 14 '12 at 16:39






  • 6




    $begingroup$
    @agent154: By the way, where is $frac13$ in your list?
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:40








  • 2




    $begingroup$
    @agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:44








  • 6




    $begingroup$
    ...are you sure you understand Cantor's diagonal argument?
    $endgroup$
    – camccann
    Nov 14 '12 at 16:45






  • 5




    $begingroup$
    @MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
    $endgroup$
    – Thomas Andrews
    Nov 14 '12 at 16:45














4












4








4


5



$begingroup$


I understand Cantor's diagonal argument, but it just doesn't seem to jive for some reason.



Lets say I assign the following numbers ad infinitum...




  1. $1to 0.1$

  2. $2to 0.2$

  3. $3to 0.3$
    ...

  4. $10to 0.10$

  5. $11to 0.11$


and so on... How come there's supposedly at least one more real number than you can map to a member of $mathbb{N}$?










share|cite|improve this question











$endgroup$




I understand Cantor's diagonal argument, but it just doesn't seem to jive for some reason.



Lets say I assign the following numbers ad infinitum...




  1. $1to 0.1$

  2. $2to 0.2$

  3. $3to 0.3$
    ...

  4. $10to 0.10$

  5. $11to 0.11$


and so on... How come there's supposedly at least one more real number than you can map to a member of $mathbb{N}$?







elementary-set-theory real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 20:04









Xander Henderson

14.4k103554




14.4k103554










asked Nov 14 '12 at 16:35









agent154agent154

3,667196398




3,667196398








  • 5




    $begingroup$
    Because... Cantor explicitly tells us what that "one" real number is... ???
    $endgroup$
    – The Chaz 2.0
    Nov 14 '12 at 16:39






  • 6




    $begingroup$
    @agent154: By the way, where is $frac13$ in your list?
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:40








  • 2




    $begingroup$
    @agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:44








  • 6




    $begingroup$
    ...are you sure you understand Cantor's diagonal argument?
    $endgroup$
    – camccann
    Nov 14 '12 at 16:45






  • 5




    $begingroup$
    @MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
    $endgroup$
    – Thomas Andrews
    Nov 14 '12 at 16:45














  • 5




    $begingroup$
    Because... Cantor explicitly tells us what that "one" real number is... ???
    $endgroup$
    – The Chaz 2.0
    Nov 14 '12 at 16:39






  • 6




    $begingroup$
    @agent154: By the way, where is $frac13$ in your list?
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:40








  • 2




    $begingroup$
    @agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
    $endgroup$
    – Dejan Govc
    Nov 14 '12 at 16:44








  • 6




    $begingroup$
    ...are you sure you understand Cantor's diagonal argument?
    $endgroup$
    – camccann
    Nov 14 '12 at 16:45






  • 5




    $begingroup$
    @MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
    $endgroup$
    – Thomas Andrews
    Nov 14 '12 at 16:45








5




5




$begingroup$
Because... Cantor explicitly tells us what that "one" real number is... ???
$endgroup$
– The Chaz 2.0
Nov 14 '12 at 16:39




$begingroup$
Because... Cantor explicitly tells us what that "one" real number is... ???
$endgroup$
– The Chaz 2.0
Nov 14 '12 at 16:39




6




6




$begingroup$
@agent154: By the way, where is $frac13$ in your list?
$endgroup$
– Dejan Govc
Nov 14 '12 at 16:40






$begingroup$
@agent154: By the way, where is $frac13$ in your list?
$endgroup$
– Dejan Govc
Nov 14 '12 at 16:40






2




2




$begingroup$
@agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
$endgroup$
– Dejan Govc
Nov 14 '12 at 16:44






$begingroup$
@agent154: Not exactly; $frac13=0.333333333ldots$ has an infinite number of 3's in its decimal expansion.
$endgroup$
– Dejan Govc
Nov 14 '12 at 16:44






6




6




$begingroup$
...are you sure you understand Cantor's diagonal argument?
$endgroup$
– camccann
Nov 14 '12 at 16:45




$begingroup$
...are you sure you understand Cantor's diagonal argument?
$endgroup$
– camccann
Nov 14 '12 at 16:45




5




5




$begingroup$
@MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
$endgroup$
– Thomas Andrews
Nov 14 '12 at 16:45




$begingroup$
@MakotoKato That's a bit of a useless argument. Presumably, if he knew what the Lebesgue measure was, he would already be comfortable with Cantor's argument.
$endgroup$
– Thomas Andrews
Nov 14 '12 at 16:45










4 Answers
4






active

oldest

votes


















9












$begingroup$

If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111ldots$ (which happens to be equal to $frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?



(If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $frac 13, frac23, frac 16, frac 17, frac 27ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
    $endgroup$
    – Golob
    Nov 14 '12 at 21:26










  • $begingroup$
    @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
    $endgroup$
    – agent154
    Nov 15 '12 at 19:41





















5












$begingroup$

Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:




  1. 0.12345...

  2. 0.32145...

  3. 0.87654...

  4. 0.28374...

  5. 0.67676...

  6. etc...


From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:



$$0.12676...$$



Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:



$$0.12676... mapsto 0.21111...$$



This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.



Check out this YouTube clip.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
    $endgroup$
    – agent154
    Nov 15 '12 at 19:44










  • $begingroup$
    @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
    $endgroup$
    – Fly by Night
    Nov 15 '12 at 20:26










  • $begingroup$
    well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
    $endgroup$
    – agent154
    Nov 16 '12 at 6:09










  • $begingroup$
    Good example. Most proofs on this are too long.
    $endgroup$
    – Aåkon
    Jan 9 '16 at 19:31



















1












$begingroup$

How come there's supposedly at least one more real number than you can map to a member of $mathbb N$?



Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers. Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence. But that contradicts our supposition! Hence our supposition - that we can have an enumeration of all reals - is false. That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.



Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript). Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list.



    First a few properties:




    • You know that two numbers differ if just one digit differs.

    • If a number shares the previous property with every number in a set, it is not part of the set.


    Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list.



    This is why Cantor's diagonal as a method proves the result that the reals are uncountable.






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9












      $begingroup$

      If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111ldots$ (which happens to be equal to $frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?



      (If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $frac 13, frac23, frac 16, frac 17, frac 27ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
        $endgroup$
        – Golob
        Nov 14 '12 at 21:26










      • $begingroup$
        @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
        $endgroup$
        – agent154
        Nov 15 '12 at 19:41


















      9












      $begingroup$

      If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111ldots$ (which happens to be equal to $frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?



      (If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $frac 13, frac23, frac 16, frac 17, frac 27ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
        $endgroup$
        – Golob
        Nov 14 '12 at 21:26










      • $begingroup$
        @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
        $endgroup$
        – agent154
        Nov 15 '12 at 19:41
















      9












      9








      9





      $begingroup$

      If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111ldots$ (which happens to be equal to $frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?



      (If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $frac 13, frac23, frac 16, frac 17, frac 27ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)






      share|cite|improve this answer











      $endgroup$



      If we apply the Cantor diagonal argument to the list you have there, and assign the $n$th digit in our constructed number to be 0 if the $n$th digit of the $n$th number in the list is 1, and 1 if it isn't, we construct the real number $0.0111111ldots$ (which happens to be equal to $frac1{90}$) and in fact is not anywhere in your list, since there is no integer that looks like $0111111ldots$. So the construction worked fine: you presented what you claim is a list of all real numbers, and then the construction produced a real number that you omitted. Where's the problem?



      (If you're not happy with that, perhaps it's enough to point out that your list not only omits all the repeating decimals such as $frac 13, frac23, frac 16, frac 17, frac 27ldots$, but also all the numbers less than 1/10. So it's not even a list of all the rational numbers.)







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Oct 9 '14 at 18:57

























      answered Nov 14 '12 at 18:24









      MJDMJD

      47.3k29213396




      47.3k29213396












      • $begingroup$
        the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
        $endgroup$
        – Golob
        Nov 14 '12 at 21:26










      • $begingroup$
        @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
        $endgroup$
        – agent154
        Nov 15 '12 at 19:41




















      • $begingroup$
        the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
        $endgroup$
        – Golob
        Nov 14 '12 at 21:26










      • $begingroup$
        @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
        $endgroup$
        – agent154
        Nov 15 '12 at 19:41


















      $begingroup$
      the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
      $endgroup$
      – Golob
      Nov 14 '12 at 21:26




      $begingroup$
      the fact that you can inject $mathbb R$ into $mathbb N$ doesn't prove that there are more real numbers then natural ones. It just shows that there are no less real numbers then natural ones.
      $endgroup$
      – Golob
      Nov 14 '12 at 21:26












      $begingroup$
      @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
      $endgroup$
      – agent154
      Nov 15 '12 at 19:41






      $begingroup$
      @mjd That actually helps make a bit more sense. At least my assignment doesn't work since it doesn't account for numbers less than 1/10 as you state. But I guess my issue is with the fact that I don't understand this well enough yet. I only just learned about countability, and even then very poorly, since my prof is not a good teacher.
      $endgroup$
      – agent154
      Nov 15 '12 at 19:41













      5












      $begingroup$

      Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:




      1. 0.12345...

      2. 0.32145...

      3. 0.87654...

      4. 0.28374...

      5. 0.67676...

      6. etc...


      From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:



      $$0.12676...$$



      Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:



      $$0.12676... mapsto 0.21111...$$



      This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.



      Check out this YouTube clip.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
        $endgroup$
        – agent154
        Nov 15 '12 at 19:44










      • $begingroup$
        @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
        $endgroup$
        – Fly by Night
        Nov 15 '12 at 20:26










      • $begingroup$
        well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
        $endgroup$
        – agent154
        Nov 16 '12 at 6:09










      • $begingroup$
        Good example. Most proofs on this are too long.
        $endgroup$
        – Aåkon
        Jan 9 '16 at 19:31
















      5












      $begingroup$

      Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:




      1. 0.12345...

      2. 0.32145...

      3. 0.87654...

      4. 0.28374...

      5. 0.67676...

      6. etc...


      From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:



      $$0.12676...$$



      Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:



      $$0.12676... mapsto 0.21111...$$



      This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.



      Check out this YouTube clip.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
        $endgroup$
        – agent154
        Nov 15 '12 at 19:44










      • $begingroup$
        @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
        $endgroup$
        – Fly by Night
        Nov 15 '12 at 20:26










      • $begingroup$
        well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
        $endgroup$
        – agent154
        Nov 16 '12 at 6:09










      • $begingroup$
        Good example. Most proofs on this are too long.
        $endgroup$
        – Aåkon
        Jan 9 '16 at 19:31














      5












      5








      5





      $begingroup$

      Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:




      1. 0.12345...

      2. 0.32145...

      3. 0.87654...

      4. 0.28374...

      5. 0.67676...

      6. etc...


      From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:



      $$0.12676...$$



      Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:



      $$0.12676... mapsto 0.21111...$$



      This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.



      Check out this YouTube clip.






      share|cite|improve this answer











      $endgroup$



      Let's look at the interval $0 < x < 1.$ Let's assume all of its numbers are countable, i.e. they are listable. Let's say the list goes like:




      1. 0.12345...

      2. 0.32145...

      3. 0.87654...

      4. 0.28374...

      5. 0.67676...

      6. etc...


      From the list, I make a new number. Its first decimal place is the same as the first number's, its second decimal place is the same as the second number's, its third decimal place is the same as the third number's, etc. The new number I get is:



      $$0.12676...$$



      Using this number, I create a new number. If a digit is a $1$ then I change it to a $2$. If a digit is anything else then I change it to a $1$. Thus:



      $$0.12676... mapsto 0.21111...$$



      This new number, $0.21111...$ is not on my list. It's not the first number because it's different in the first place, it's not the second number because it's different in the second place, it's not the third number because it's different in the third place, it's not the fourth number because it's different in the fourth place, etc. It follows that my new number is not in my list, and so I could not have listed all of the numbers $0 < x < 1$. Thus, by contradiction, the interval is not countable.



      Check out this YouTube clip.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 14 '12 at 19:30

























      answered Nov 14 '12 at 19:24









      Fly by NightFly by Night

      25.9k32978




      25.9k32978












      • $begingroup$
        As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
        $endgroup$
        – agent154
        Nov 15 '12 at 19:44










      • $begingroup$
        @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
        $endgroup$
        – Fly by Night
        Nov 15 '12 at 20:26










      • $begingroup$
        well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
        $endgroup$
        – agent154
        Nov 16 '12 at 6:09










      • $begingroup$
        Good example. Most proofs on this are too long.
        $endgroup$
        – Aåkon
        Jan 9 '16 at 19:31


















      • $begingroup$
        As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
        $endgroup$
        – agent154
        Nov 15 '12 at 19:44










      • $begingroup$
        @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
        $endgroup$
        – Fly by Night
        Nov 15 '12 at 20:26










      • $begingroup$
        well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
        $endgroup$
        – agent154
        Nov 16 '12 at 6:09










      • $begingroup$
        Good example. Most proofs on this are too long.
        $endgroup$
        – Aåkon
        Jan 9 '16 at 19:31
















      $begingroup$
      As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
      $endgroup$
      – agent154
      Nov 15 '12 at 19:44




      $begingroup$
      As stated, I do understand the basic theory of the diagonalization method... but despite it saying that the new number couldn't possibly exist in the list, I just couldn't see a logical reason why I couldn't enumerate it as stated above. But I hadn't noticed that doing it my way at the very least omits all the numbers less than 1/10
      $endgroup$
      – agent154
      Nov 15 '12 at 19:44












      $begingroup$
      @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
      $endgroup$
      – Fly by Night
      Nov 15 '12 at 20:26




      $begingroup$
      @agent154 To be blunt: you obviously didn't understand the basic theory. If you had then you wouldn't have posted the question that you did.
      $endgroup$
      – Fly by Night
      Nov 15 '12 at 20:26












      $begingroup$
      well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
      $endgroup$
      – agent154
      Nov 16 '12 at 6:09




      $begingroup$
      well that's not fair. Many smart people claim to have some kind of proof that Cantor was wrong, but it doesn't mean they don't understand his theory. I just can't take something at face value sometimes. I understand the logic behind the diagonal argument, but neglecting that, I just can't wrap my head around why you cannot possibly enumerate all the real numbers.
      $endgroup$
      – agent154
      Nov 16 '12 at 6:09












      $begingroup$
      Good example. Most proofs on this are too long.
      $endgroup$
      – Aåkon
      Jan 9 '16 at 19:31




      $begingroup$
      Good example. Most proofs on this are too long.
      $endgroup$
      – Aåkon
      Jan 9 '16 at 19:31











      1












      $begingroup$

      How come there's supposedly at least one more real number than you can map to a member of $mathbb N$?



      Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers. Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence. But that contradicts our supposition! Hence our supposition - that we can have an enumeration of all reals - is false. That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.



      Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript). Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        How come there's supposedly at least one more real number than you can map to a member of $mathbb N$?



        Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers. Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence. But that contradicts our supposition! Hence our supposition - that we can have an enumeration of all reals - is false. That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.



        Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript). Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          How come there's supposedly at least one more real number than you can map to a member of $mathbb N$?



          Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers. Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence. But that contradicts our supposition! Hence our supposition - that we can have an enumeration of all reals - is false. That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.



          Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript). Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.






          share|cite|improve this answer









          $endgroup$



          How come there's supposedly at least one more real number than you can map to a member of $mathbb N$?



          Well, suppose there isn't - that Cantor's conclusion, his theorem, is wrong, because our enumeration covers all real numbers. Wonderful, but let us see what happens when we take our enumeration and apply Cantor's diagonal technique to obtain a real number that can't be in this sequence. But that contradicts our supposition! Hence our supposition - that we can have an enumeration of all reals - is false. That the argument can be applied to any enumeration is what it takes for Cantor's theorem to be true.



          Wilfred Hodges wrote an excellent survey of wrong refutations of Catonr' argument, his An Editor Recalls Some Hopeless Papers (Postscript). Section 7, dealing with how the counterfactual assumption confuses, might be of particular interest.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 14 '12 at 21:12









          Charles StewartCharles Stewart

          3,3542229




          3,3542229























              0












              $begingroup$

              Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list.



              First a few properties:




              • You know that two numbers differ if just one digit differs.

              • If a number shares the previous property with every number in a set, it is not part of the set.


              Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list.



              This is why Cantor's diagonal as a method proves the result that the reals are uncountable.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list.



                First a few properties:




                • You know that two numbers differ if just one digit differs.

                • If a number shares the previous property with every number in a set, it is not part of the set.


                Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list.



                This is why Cantor's diagonal as a method proves the result that the reals are uncountable.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list.



                  First a few properties:




                  • You know that two numbers differ if just one digit differs.

                  • If a number shares the previous property with every number in a set, it is not part of the set.


                  Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list.



                  This is why Cantor's diagonal as a method proves the result that the reals are uncountable.






                  share|cite|improve this answer









                  $endgroup$



                  Cantor's diagonal is a trick to show that given any list of reals, a real can be found that is not in the list.



                  First a few properties:




                  • You know that two numbers differ if just one digit differs.

                  • If a number shares the previous property with every number in a set, it is not part of the set.


                  Cantor's diagonal is a clever solution to finding a number which satisfies these properties. The number which is the diagonal is transformed s.t. it doesn't share the first digit of the first number nor the second digit with the second and so on. Thus the number is unique to the list.



                  This is why Cantor's diagonal as a method proves the result that the reals are uncountable.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 6 '15 at 21:08









                  cdosborncdosborn

                  1093




                  1093






























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