Proof that R is uncountable using diagonalization The Next CEO of Stack OverflowWhy does Cantor's Proof (that R is uncountable) fail for Q?Infinite Cartesian product of countable sets is uncountableis it possible to proof that this number is not rationalProof that $Bbb R$ is uncountable.show $[0,1]$ is uncountable using outer measureHow to pick decimal expansion in the proof that $(0,1)$ uncountableUncountable intersection of uncountable setsFalse proofs claiming that $mathbbQ$ is uncountable.Proving that the interval $(0,1)$ is uncountableGeneral methods for proving countability and uncountability
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Proof that R is uncountable using diagonalization
The Next CEO of Stack OverflowWhy does Cantor's Proof (that R is uncountable) fail for Q?Infinite Cartesian product of countable sets is uncountableis it possible to proof that this number is not rationalProof that $Bbb R$ is uncountable.show $[0,1]$ is uncountable using outer measureHow to pick decimal expansion in the proof that $(0,1)$ uncountableUncountable intersection of uncountable setsFalse proofs claiming that $mathbbQ$ is uncountable.Proving that the interval $(0,1)$ is uncountableGeneral methods for proving countability and uncountability
$begingroup$
I understand that in order to prove that $BbbR$ is uncountable, one must show that the following list can be compiled:
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
But from here I am not sure how one might show that there is some decimal expansion in $(0,1)$ that is not ‘hit’, and so that $f:BbbNrightarrow(0,1)$ is not bijective, and so that $BbbR$ is not countable.
real-analysis analysis
asked Mar 27 at 18:55
Ali LodhiAli Lodhi
343
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.
$endgroup$
add a comment |
$begingroup$
I understand that in order to prove that $BbbR$ is uncountable, one must show that the following list can be compiled:
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
But from here I am not sure how one might show that there is some decimal expansion in $(0,1)$ that is not ‘hit’, and so that $f:BbbNrightarrow(0,1)$ is not bijective, and so that $BbbR$ is not countable.
real-analysis analysis
asked Mar 27 at 18:55
Ali LodhiAli Lodhi
343
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.
$endgroup$
2
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13
add a comment |
$begingroup$
I understand that in order to prove that $BbbR$ is uncountable, one must show that the following list can be compiled:
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
But from here I am not sure how one might show that there is some decimal expansion in $(0,1)$ that is not ‘hit’, and so that $f:BbbNrightarrow(0,1)$ is not bijective, and so that $BbbR$ is not countable.
real-analysis analysis
asked Mar 27 at 18:55
Ali LodhiAli Lodhi
343
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.
$endgroup$
I understand that in order to prove that $BbbR$ is uncountable, one must show that the following list can be compiled:
$$x_1=x_11x_12x_13\x_2=x_21x_22x_23\x_3=x_31x_32x_33\.\.\.$$
But from here I am not sure how one might show that there is some decimal expansion in $(0,1)$ that is not ‘hit’, and so that $f:BbbNrightarrow(0,1)$ is not bijective, and so that $BbbR$ is not countable.
real-analysis analysis
real-analysis analysis
asked Mar 27 at 18:55
Ali LodhiAli Lodhi
343
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 27 at 18:55
Ali LodhiAli Lodhi
343
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 27 at 18:55
Ali LodhiAli Lodhi
343
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 27 at 18:55
Ali LodhiAli Lodhi
343
asked Mar 27 at 18:55
Ali LodhiAli Lodhi
343
343
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.
Ali Lodhi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13
add a comment |
2
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13
2
2
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13
add a comment |
1 Answer
1
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$begingroup$
Consider a sequence $y=y_1y_2y_3...$ such that for each $i$, $y_i neq x_ii$.
If $y=x_n$ for an integer $n$, then considering the $n-$th decimal, you would have $y_n=x_nn$. This is not true.
So $y$ is not one of the $x_n$.
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
$endgroup$
add a comment |
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$begingroup$
Consider a sequence $y=y_1y_2y_3...$ such that for each $i$, $y_i neq x_ii$.
If $y=x_n$ for an integer $n$, then considering the $n-$th decimal, you would have $y_n=x_nn$. This is not true.
So $y$ is not one of the $x_n$.
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
$endgroup$
add a comment |
$begingroup$
Consider a sequence $y=y_1y_2y_3...$ such that for each $i$, $y_i neq x_ii$.
If $y=x_n$ for an integer $n$, then considering the $n-$th decimal, you would have $y_n=x_nn$. This is not true.
So $y$ is not one of the $x_n$.
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
$endgroup$
add a comment |
$begingroup$
Consider a sequence $y=y_1y_2y_3...$ such that for each $i$, $y_i neq x_ii$.
If $y=x_n$ for an integer $n$, then considering the $n-$th decimal, you would have $y_n=x_nn$. This is not true.
So $y$ is not one of the $x_n$.
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
$endgroup$
Consider a sequence $y=y_1y_2y_3...$ such that for each $i$, $y_i neq x_ii$.
If $y=x_n$ for an integer $n$, then considering the $n-$th decimal, you would have $y_n=x_nn$. This is not true.
So $y$ is not one of the $x_n$.
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
answered Mar 27 at 19:08
TheSilverDoeTheSilverDoe
4,884215
4,884215
add a comment |
add a comment |
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.
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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2
$begingroup$
You need to show the list $cannot$ be compiled. The argument is by contradiction: Assume it is countable, therefore it can be "listed". Now you assume that is an exhaustive list of its elements, and then create a new element not contained in the list. Therefore the list is not exhaustive. Now, since adding that new element to the list does not solve the problem, the original assumption of countability was wrong
$endgroup$
– Fede Poncio
Mar 27 at 19:08
$begingroup$
Precisely, @FedePoncio. And you have given a good concise explanation of the proof for OP.
$endgroup$
– Lubin
Mar 27 at 19:13