Mixing
Why can’t you reassign the ‘mystery number’ in Cantor’s diagonal argument to a new number in the...
$begingroup$
I don’t want to claim that I have ‘refuted Cantor’ or something here, I just want to understand it adequately.
I do understand that the proof works something like this:
You assume that you can map the naturals onto the real numbers like so, where each letter represents some arbitrary digit:
$$1 — 0.abcdcdots$$
$$2 — 0.efghcdots$$
$$3 — 0.ijklcdots$$
And then you constrict the ‘mystery number’ $x$ which differs in at least one digit of each number that is assigned to a natural number. This mystery number, by definition, cannot be mapped onto any of the natural numbers. This is fine, but why can’t you just assign $0.abcdcdots$ to $2, 0.efghcdots$ to $3$, and so on, then assign $x$ to $1$, which would now be open?
What is the problem with doing this?
Update: Sorry for the poor notation and formatting, I’m writing this on my phone.
elementary-set-theory infinity
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
$endgroup$
add a comment |
$begingroup$
I don’t want to claim that I have ‘refuted Cantor’ or something here, I just want to understand it adequately.
I do understand that the proof works something like this:
You assume that you can map the naturals onto the real numbers like so, where each letter represents some arbitrary digit:
$$1 — 0.abcdcdots$$
$$2 — 0.efghcdots$$
$$3 — 0.ijklcdots$$
And then you constrict the ‘mystery number’ $x$ which differs in at least one digit of each number that is assigned to a natural number. This mystery number, by definition, cannot be mapped onto any of the natural numbers. This is fine, but why can’t you just assign $0.abcdcdots$ to $2, 0.efghcdots$ to $3$, and so on, then assign $x$ to $1$, which would now be open?
What is the problem with doing this?
Update: Sorry for the poor notation and formatting, I’m writing this on my phone.
elementary-set-theory infinity
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
$endgroup$
9
$begingroup$
Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
$endgroup$
– Cheerful Parsnip
Jan 18 at 2:57
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
5
$begingroup$
Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
$endgroup$
– spaceisdarkgreen
Jan 18 at 3:12
1
$begingroup$
The diagonal argument doesn't assert that only one number is missing from the list.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:53
$begingroup$
The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
$endgroup$
– Lee Mosher
Jan 20 at 18:28
add a comment |
$begingroup$
I don’t want to claim that I have ‘refuted Cantor’ or something here, I just want to understand it adequately.
I do understand that the proof works something like this:
You assume that you can map the naturals onto the real numbers like so, where each letter represents some arbitrary digit:
$$1 — 0.abcdcdots$$
$$2 — 0.efghcdots$$
$$3 — 0.ijklcdots$$
And then you constrict the ‘mystery number’ $x$ which differs in at least one digit of each number that is assigned to a natural number. This mystery number, by definition, cannot be mapped onto any of the natural numbers. This is fine, but why can’t you just assign $0.abcdcdots$ to $2, 0.efghcdots$ to $3$, and so on, then assign $x$ to $1$, which would now be open?
What is the problem with doing this?
Update: Sorry for the poor notation and formatting, I’m writing this on my phone.
elementary-set-theory infinity
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
$endgroup$
I don’t want to claim that I have ‘refuted Cantor’ or something here, I just want to understand it adequately.
I do understand that the proof works something like this:
You assume that you can map the naturals onto the real numbers like so, where each letter represents some arbitrary digit:
$$1 — 0.abcdcdots$$
$$2 — 0.efghcdots$$
$$3 — 0.ijklcdots$$
And then you constrict the ‘mystery number’ $x$ which differs in at least one digit of each number that is assigned to a natural number. This mystery number, by definition, cannot be mapped onto any of the natural numbers. This is fine, but why can’t you just assign $0.abcdcdots$ to $2, 0.efghcdots$ to $3$, and so on, then assign $x$ to $1$, which would now be open?
What is the problem with doing this?
Update: Sorry for the poor notation and formatting, I’m writing this on my phone.
elementary-set-theory infinity
elementary-set-theory infinity
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
edited Jan 18 at 3:45
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
asked Jan 18 at 2:52
Caleb MahlenCaleb Mahlen
262
262
9
$begingroup$
Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
$endgroup$
– Cheerful Parsnip
Jan 18 at 2:57
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
5
$begingroup$
Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
$endgroup$
– spaceisdarkgreen
Jan 18 at 3:12
1
$begingroup$
The diagonal argument doesn't assert that only one number is missing from the list.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:53
$begingroup$
The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
$endgroup$
– Lee Mosher
Jan 20 at 18:28
add a comment |
9
$begingroup$
Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
$endgroup$
– Cheerful Parsnip
Jan 18 at 2:57
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
5
$begingroup$
Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
$endgroup$
– spaceisdarkgreen
Jan 18 at 3:12
1
$begingroup$
The diagonal argument doesn't assert that only one number is missing from the list.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:53
$begingroup$
The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
$endgroup$
– Lee Mosher
Jan 20 at 18:28
9
9
$begingroup$
Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
$endgroup$
– Cheerful Parsnip
Jan 18 at 2:57
$begingroup$
Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
$endgroup$
– Cheerful Parsnip
Jan 18 at 2:57
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
$begingroup$
Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
$endgroup$
– dantopa
Jan 18 at 3:00
5
5
$begingroup$
Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
$endgroup$
– spaceisdarkgreen
Jan 18 at 3:12
$begingroup$
Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
$endgroup$
– spaceisdarkgreen
Jan 18 at 3:12
1
1
$begingroup$
The diagonal argument doesn't assert that only one number is missing from the list.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:53
$begingroup$
The diagonal argument doesn't assert that only one number is missing from the list.
$endgroup$
– Lord Shark the Unknown
Jan 18 at 4:53
$begingroup$
The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
$endgroup$
– Lee Mosher
Jan 20 at 18:28
$begingroup$
The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
$endgroup$
– Lee Mosher
Jan 20 at 18:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
That is actually a valid question, but it is based on a misunderstanding of Cantor's Diagonal Proof. That is not your fault, it is taught incorrectly; usually something like this outline:
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces every element t in T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- (A) and (B) together prove that f(n) does not produce every element t in T.
- Since #4 contradicts the assumption in #2, that assumption must be false.
The problem with this, is that to use Proof by Contradiction, you have to use every part of what you assumed to produce the contradiction. Any part you don't use isn't proven by the contradiction. All step #3 really uses, is that every f(n) is an element of T, not that every element of T is produced. So wondering if you can put it back in is a valid question. You can't, but to see why it helps to use Cantor's actual outline.
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces elements of T.
- The set of all elements produced by f(n) is the set S. Nothing is assumed at this point about whether S=T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- The two statement S=T and "There is a t0 that must be in T, but is different than every f(n)" can't both be true. Since we know that the second one is true, the first can't be.
- You can't count every element of T.
answered Jan 20 at 18:24
JeffJoJeffJo
1654
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
$endgroup$
add a comment |
$begingroup$
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
$endgroup$
Yes, if you are missing a real number, then you can always add it to the front of your list, but this not mean that the new list is now complete. (this is LordSharktheUnknown's point: if the mystery number was the only one that as missing, then yes, you'd get a complete list, but it may not be the only one missing).
In fact, we know the new list cannot be complete, because we can go through the same diagonalization to find a new mystery number that is not on this new list (this is spaceisdarkgreen's point)
And finally, remember that when you do the diagonalization on the original list, we did this within a proof by contradiction. That is, we assumed that there is a complete list ... but when we take any such supposedly complete list we can find a mystery number that is not on the list ... and so we reach a contradiction already. Coming up with a new list does not take away from the contradiction that you obtain when assuming there is a complete list at all, so you are doing nothing to refute the proof (and this is Cheerful Parsnip's point)
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
edited Jan 20 at 18:41
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
answered Jan 19 at 16:31
Bram28Bram28
63.2k44793
63.2k44793
add a comment |
add a comment |
$begingroup$
That is actually a valid question, but it is based on a misunderstanding of Cantor's Diagonal Proof. That is not your fault, it is taught incorrectly; usually something like this outline:
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces every element t in T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- (A) and (B) together prove that f(n) does not produce every element t in T.
- Since #4 contradicts the assumption in #2, that assumption must be false.
The problem with this, is that to use Proof by Contradiction, you have to use every part of what you assumed to produce the contradiction. Any part you don't use isn't proven by the contradiction. All step #3 really uses, is that every f(n) is an element of T, not that every element of T is produced. So wondering if you can put it back in is a valid question. You can't, but to see why it helps to use Cantor's actual outline.
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces elements of T.
- The set of all elements produced by f(n) is the set S. Nothing is assumed at this point about whether S=T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- The two statement S=T and "There is a t0 that must be in T, but is different than every f(n)" can't both be true. Since we know that the second one is true, the first can't be.
- You can't count every element of T.
answered Jan 20 at 18:24
JeffJoJeffJo
1654
$endgroup$
add a comment |
$begingroup$
That is actually a valid question, but it is based on a misunderstanding of Cantor's Diagonal Proof. That is not your fault, it is taught incorrectly; usually something like this outline:
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces every element t in T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- (A) and (B) together prove that f(n) does not produce every element t in T.
- Since #4 contradicts the assumption in #2, that assumption must be false.
The problem with this, is that to use Proof by Contradiction, you have to use every part of what you assumed to produce the contradiction. Any part you don't use isn't proven by the contradiction. All step #3 really uses, is that every f(n) is an element of T, not that every element of T is produced. So wondering if you can put it back in is a valid question. You can't, but to see why it helps to use Cantor's actual outline.
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces elements of T.
- The set of all elements produced by f(n) is the set S. Nothing is assumed at this point about whether S=T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- The two statement S=T and "There is a t0 that must be in T, but is different than every f(n)" can't both be true. Since we know that the second one is true, the first can't be.
- You can't count every element of T.
answered Jan 20 at 18:24
JeffJoJeffJo
1654
$endgroup$
add a comment |
$begingroup$
That is actually a valid question, but it is based on a misunderstanding of Cantor's Diagonal Proof. That is not your fault, it is taught incorrectly; usually something like this outline:
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces every element t in T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- (A) and (B) together prove that f(n) does not produce every element t in T.
- Since #4 contradicts the assumption in #2, that assumption must be false.
The problem with this, is that to use Proof by Contradiction, you have to use every part of what you assumed to produce the contradiction. Any part you don't use isn't proven by the contradiction. All step #3 really uses, is that every f(n) is an element of T, not that every element of T is produced. So wondering if you can put it back in is a valid question. You can't, but to see why it helps to use Cantor's actual outline.
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces elements of T.
- The set of all elements produced by f(n) is the set S. Nothing is assumed at this point about whether S=T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- The two statement S=T and "There is a t0 that must be in T, but is different than every f(n)" can't both be true. Since we know that the second one is true, the first can't be.
- You can't count every element of T.
answered Jan 20 at 18:24
JeffJoJeffJo
1654
$endgroup$
That is actually a valid question, but it is based on a misunderstanding of Cantor's Diagonal Proof. That is not your fault, it is taught incorrectly; usually something like this outline:
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces every element t in T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- (A) and (B) together prove that f(n) does not produce every element t in T.
- Since #4 contradicts the assumption in #2, that assumption must be false.
The problem with this, is that to use Proof by Contradiction, you have to use every part of what you assumed to produce the contradiction. Any part you don't use isn't proven by the contradiction. All step #3 really uses, is that every f(n) is an element of T, not that every element of T is produced. So wondering if you can put it back in is a valid question. You can't, but to see why it helps to use Cantor's actual outline.
- For some infinite set T:
- Assume there is a mapping f(n), for n=1,2,3,..., that produces elements of T.
- The set of all elements produced by f(n) is the set S. Nothing is assumed at this point about whether S=T.
- Use Diagonalization to construct a new element t(0) that (A) must be in T, but (B) is different than every f(n).
- The two statement S=T and "There is a t0 that must be in T, but is different than every f(n)" can't both be true. Since we know that the second one is true, the first can't be.
- You can't count every element of T.
answered Jan 20 at 18:24
JeffJoJeffJo
1654
answered Jan 20 at 18:24
JeffJoJeffJo
1654
answered Jan 20 at 18:24
JeffJoJeffJo
1654
answered Jan 20 at 18:24
JeffJoJeffJo
1654
1654
add a comment |
add a comment |
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9
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Sure you can accommodate a missing number on the list, but according to your assumption it was already there. That's the contradiction.
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– Cheerful Parsnip
Jan 18 at 2:57
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Welcome to Mathematics Stack Exchange. Take the short tour to see how how to get the most from your time here. For typesetting equations we use MathJax.
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– dantopa
Jan 18 at 3:00
5
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Sure, you can just add the real you constructed to the front of the list, but then you can construct another real, by the same method, that is not on your new list. The statement is "for any list of real numbers, there is a real not on that list" (i.e. there is no surjection $mathbb Nto mathbb R$). That is exactly what the diagonal argument shows, and your observation does nothing to refute that.
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– spaceisdarkgreen
Jan 18 at 3:12
1
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The diagonal argument doesn't assert that only one number is missing from the list.
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– Lord Shark the Unknown
Jan 18 at 4:53
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The diagonal argument can be summarized like this: no matter what you do, no matter how you make your list, the list will be incomplete.
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– Lee Mosher
Jan 20 at 18:28