Proving that R is uncountable The Next CEO of Stack OverflowIs there a bijective map from $(0,1)$ to $mathbbR$?Cantor's Diagonalization: Impossible to formulate the decimal expansion in (0, 1) that serves as the contradiction?Why does Cantor's Proof (that R is uncountable) fail for Q?is it possible to proof that this number is not rationalProof that $Bbb R$ is uncountable.Proof of the uncountability of reals using the diagonal argument—problem?How to pick decimal expansion in the proof that $(0,1)$ uncountableDifference in way of proving [0,1] is uncountable, (0,1) is uncountable, and etcFalse proofs claiming that $mathbbQ$ is uncountable.Proving that the interval $(0,1)$ is uncountableProof that R is uncountable using diagonalization
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Proving that R is uncountable
The Next CEO of Stack OverflowIs there a bijective map from $(0,1)$ to $mathbbR$?Cantor's Diagonalization: Impossible to formulate the decimal expansion in (0, 1) that serves as the contradiction?Why does Cantor's Proof (that R is uncountable) fail for Q?is it possible to proof that this number is not rationalProof that $Bbb R$ is uncountable.Proof of the uncountability of reals using the diagonal argument—problem?How to pick decimal expansion in the proof that $(0,1)$ uncountableDifference in way of proving [0,1] is uncountable, (0,1) is uncountable, and etcFalse proofs claiming that $mathbbQ$ is uncountable.Proving that the interval $(0,1)$ is uncountableProof that R is uncountable using diagonalization
$begingroup$
Is the following proof for the uncountability of $BbbR$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:BbbNrightarrow(0,1)$
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
Where $x_ij$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_j=x_ii+1$ when $x_iineq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_1$ in the first decimal, $x_2$ in the second, and so on. So $f:BbbNrightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $BbbR$.
real-analysis analysis
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Is the following proof for the uncountability of $BbbR$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:BbbNrightarrow(0,1)$
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
Where $x_ij$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_j=x_ii+1$ when $x_iineq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_1$ in the first decimal, $x_2$ in the second, and so on. So $f:BbbNrightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $BbbR$.
real-analysis analysis
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14
add a comment |
$begingroup$
Is the following proof for the uncountability of $BbbR$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:BbbNrightarrow(0,1)$
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
Where $x_ij$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_j=x_ii+1$ when $x_iineq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_1$ in the first decimal, $x_2$ in the second, and so on. So $f:BbbNrightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $BbbR$.
real-analysis analysis
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
Is the following proof for the uncountability of $BbbR$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:BbbNrightarrow(0,1)$
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
Where $x_ij$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_j=x_ii+1$ when $x_iineq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_1$ in the first decimal, $x_2$ in the second, and so on. So $f:BbbNrightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $BbbR$.
real-analysis analysis
real-analysis analysis
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
asked Mar 28 at 9:10
Ali LodhiAli Lodhi
383
383
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
1
$begingroup$
As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14
add a comment |
1
$begingroup$
As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14
1
1
$begingroup$
As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14
$begingroup$
As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3ldots$
But this is wrong. Let's say $x_1 = frac12,$ but you choose the unusual representation $0.49999ldots$ for it. Now it happens by accident that $x_22=x_33=x_44=ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.
Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.
If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
$endgroup$
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
add a comment |
$begingroup$
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.
The binary version goes like this:
Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:
$$a_n mapsto sum_j=1^infty fraca_j2^j.$$
Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $mathbbR$ must be uncountable.
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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votes
$begingroup$
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3ldots$
But this is wrong. Let's say $x_1 = frac12,$ but you choose the unusual representation $0.49999ldots$ for it. Now it happens by accident that $x_22=x_33=x_44=ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.
Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.
If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
$endgroup$
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
add a comment |
$begingroup$
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3ldots$
But this is wrong. Let's say $x_1 = frac12,$ but you choose the unusual representation $0.49999ldots$ for it. Now it happens by accident that $x_22=x_33=x_44=ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.
Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.
If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
$endgroup$
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
add a comment |
$begingroup$
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3ldots$
But this is wrong. Let's say $x_1 = frac12,$ but you choose the unusual representation $0.49999ldots$ for it. Now it happens by accident that $x_22=x_33=x_44=ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.
Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.
If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
$endgroup$
The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3ldots$
But this is wrong. Let's say $x_1 = frac12,$ but you choose the unusual representation $0.49999ldots$ for it. Now it happens by accident that $x_22=x_33=x_44=ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.
Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.
If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
answered Mar 28 at 10:06
Reinhard MeierReinhard Meier
2,967310
2,967310
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
add a comment |
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
+1 for this. But the problem is easy to fix $-$ see my comment to the OP.
$endgroup$
– TonyK
Mar 28 at 10:16
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3....
$endgroup$
– Ali Lodhi
Mar 28 at 15:51
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
$begingroup$
@AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers.
$endgroup$
– Reinhard Meier
Mar 28 at 16:52
add a comment |
$begingroup$
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.
The binary version goes like this:
Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:
$$a_n mapsto sum_j=1^infty fraca_j2^j.$$
Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $mathbbR$ must be uncountable.
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
add a comment |
$begingroup$
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.
The binary version goes like this:
Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:
$$a_n mapsto sum_j=1^infty fraca_j2^j.$$
Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $mathbbR$ must be uncountable.
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
$endgroup$
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
add a comment |
$begingroup$
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.
The binary version goes like this:
Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:
$$a_n mapsto sum_j=1^infty fraca_j2^j.$$
Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $mathbbR$ must be uncountable.
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
$endgroup$
You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.
The binary version goes like this:
Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:
$$a_n mapsto sum_j=1^infty fraca_j2^j.$$
Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $mathbbR$ must be uncountable.
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
answered Mar 28 at 10:32
MatthewPeterMatthewPeter
191
191
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
add a comment |
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach.
$endgroup$
– Reinhard Meier
Mar 28 at 10:35
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
$begingroup$
Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity.
$endgroup$
– MatthewPeter
Mar 28 at 10:42
add a comment |
Ali Lodhi is a new contributor. Be nice, and check out our Code of Conduct.
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Ali Lodhi is a new contributor. Be nice, and check out our Code of Conduct.
Ali Lodhi is a new contributor. Be nice, and check out our Code of Conduct.
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As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_ii+5$ if $x_iile 4$, and $y_i=x_ii-5$ if $x_iige 5$. Then $y_i$ and $x_ii$ can't represent the same number.
$endgroup$
– TonyK
Mar 28 at 10:14