Proving a set to be countable
A set $S = leftlbrace left( x, y right) vert x^2 + y^2 = dfrac{1}{n^2}, text{ where } n in mathbb{N} text{ and either } x in mathbb{Q} text{ or } y in mathbb{Q} rightrbrace$ is given. I need to prove that this is countable.
I have tried looking for a bijection $f: mathbb{Q} rightarrow S$ as $f left( x right) = left( x, y right)$ where $y$ is a fixed real number such that the property of the set is satisfied.
Clearly, this function is not even a surjection.
How should we define our bijection so that we prove $S$ is countable?
real-analysis elementary-set-theory
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
add a comment |
A set $S = leftlbrace left( x, y right) vert x^2 + y^2 = dfrac{1}{n^2}, text{ where } n in mathbb{N} text{ and either } x in mathbb{Q} text{ or } y in mathbb{Q} rightrbrace$ is given. I need to prove that this is countable.
I have tried looking for a bijection $f: mathbb{Q} rightarrow S$ as $f left( x right) = left( x, y right)$ where $y$ is a fixed real number such that the property of the set is satisfied.
Clearly, this function is not even a surjection.
How should we define our bijection so that we prove $S$ is countable?
real-analysis elementary-set-theory
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
add a comment |
A set $S = leftlbrace left( x, y right) vert x^2 + y^2 = dfrac{1}{n^2}, text{ where } n in mathbb{N} text{ and either } x in mathbb{Q} text{ or } y in mathbb{Q} rightrbrace$ is given. I need to prove that this is countable.
I have tried looking for a bijection $f: mathbb{Q} rightarrow S$ as $f left( x right) = left( x, y right)$ where $y$ is a fixed real number such that the property of the set is satisfied.
Clearly, this function is not even a surjection.
How should we define our bijection so that we prove $S$ is countable?
real-analysis elementary-set-theory
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
A set $S = leftlbrace left( x, y right) vert x^2 + y^2 = dfrac{1}{n^2}, text{ where } n in mathbb{N} text{ and either } x in mathbb{Q} text{ or } y in mathbb{Q} rightrbrace$ is given. I need to prove that this is countable.
I have tried looking for a bijection $f: mathbb{Q} rightarrow S$ as $f left( x right) = left( x, y right)$ where $y$ is a fixed real number such that the property of the set is satisfied.
Clearly, this function is not even a surjection.
How should we define our bijection so that we prove $S$ is countable?
real-analysis elementary-set-theory
real-analysis elementary-set-theory
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
asked Dec 14 '18 at 4:37
Aniruddha DeshmukhAniruddha Deshmukh
934418
934418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1{n^2},tfrac1{n^2}right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{left(x,sqrt{tfrac1{n^2}-x^2}right)right}cupleft{left(x,-sqrt{tfrac1{n^2}-x^2}right)right}
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{left(sqrt{tfrac1{n^2}-y^2},yright)right}cupleft{left(-sqrt{tfrac1{n^2}-y^2},yright)right}
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes{1,2})^2$. And this last set is countable, so $S$ is countable.
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.
edited Dec 14 '18 at 12:15
Mutantoe
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
|
show 1 more comment
Note that $x^2=frac1{n^2}-y^2$ and $y^2=frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
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2 Answers
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active
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2 Answers
2
active
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oldest
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If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1{n^2},tfrac1{n^2}right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{left(x,sqrt{tfrac1{n^2}-x^2}right)right}cupleft{left(x,-sqrt{tfrac1{n^2}-x^2}right)right}
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{left(sqrt{tfrac1{n^2}-y^2},yright)right}cupleft{left(-sqrt{tfrac1{n^2}-y^2},yright)right}
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes{1,2})^2$. And this last set is countable, so $S$ is countable.
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.
edited Dec 14 '18 at 12:15
Mutantoe
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
|
show 1 more comment
If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1{n^2},tfrac1{n^2}right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{left(x,sqrt{tfrac1{n^2}-x^2}right)right}cupleft{left(x,-sqrt{tfrac1{n^2}-x^2}right)right}
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{left(sqrt{tfrac1{n^2}-y^2},yright)right}cupleft{left(-sqrt{tfrac1{n^2}-y^2},yright)right}
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes{1,2})^2$. And this last set is countable, so $S$ is countable.
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.
edited Dec 14 '18 at 12:15
Mutantoe
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
|
show 1 more comment
If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1{n^2},tfrac1{n^2}right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{left(x,sqrt{tfrac1{n^2}-x^2}right)right}cupleft{left(x,-sqrt{tfrac1{n^2}-x^2}right)right}
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{left(sqrt{tfrac1{n^2}-y^2},yright)right}cupleft{left(-sqrt{tfrac1{n^2}-y^2},yright)right}
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes{1,2})^2$. And this last set is countable, so $S$ is countable.
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.
edited Dec 14 '18 at 12:15
Mutantoe
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
If you index on $x$ and $n$, you can do the following. Let
$$
E_n=mathbb Qcap left[-tfrac1{n^2},tfrac1{n^2}right].
$$
This sets $E_n$ are countable, because they are subsets of $mathbb Q$.
Then $S=S_1cup S_2$, where
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{left(x,sqrt{tfrac1{n^2}-x^2}right)right}cupleft{left(x,-sqrt{tfrac1{n^2}-x^2}right)right}
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{left(sqrt{tfrac1{n^2}-y^2},yright)right}cupleft{left(-sqrt{tfrac1{n^2}-y^2},yright)right}
$$
We can map this to a subset of $(mathbb Ntimesmathbb Qtimes{1,2})^2$. And this last set is countable, so $S$ is countable.
The key fact is that subsets of countable sets are countable, and that finite cartesian products of countable sets are countable.
edited Dec 14 '18 at 12:15
Mutantoe
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
edited Dec 14 '18 at 12:15
Mutantoe
572412
edited Dec 14 '18 at 12:15
Mutantoe
572412
edited Dec 14 '18 at 12:15
Mutantoe
572412
572412
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
answered Dec 14 '18 at 4:50
Martin ArgeramiMartin Argerami
124k1176175
124k1176175
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
|
show 1 more comment
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
The way you have written the union for $S$, only takes into consideration when $x$ is rational and $y$ can be anything. What if it goes the other way round?
– Aniruddha Deshmukh
Dec 14 '18 at 4:52
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Or do you want to say that the final mapping would be $left( x, y right) mapsto left( n, x, 1 right)$ if $x$ is rational and $left( x, y right) mapsto left( n, y, 2 right)$ if $y$ is rational but $x$ is not.
– Aniruddha Deshmukh
Dec 14 '18 at 4:53
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
Dec 14 '18 at 4:58
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
Why are we mapping it to $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)^2$? I think $left( mathbb{N} times mathbb{Q} times leftlbrace 1, 2 rightrbrace right)$ should work with the mapping I defined in the comment.
– Aniruddha Deshmukh
Dec 14 '18 at 5:00
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
I'm using the 1 and 2 to count the positive and negative roots. There is some overlapping, but it's not a big deal. The maps as you say also work, I think.
– Martin Argerami
Dec 14 '18 at 5:04
|
show 1 more comment
Note that $x^2=frac1{n^2}-y^2$ and $y^2=frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
Note that $x^2=frac1{n^2}-y^2$ and $y^2=frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
Note that $x^2=frac1{n^2}-y^2$ and $y^2=frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
Note that $x^2=frac1{n^2}-y^2$ and $y^2=frac1{n^2}-x^2$. So if $y$ is rational, then $x^2$ is rational as well, and likewise for $y^2$ if $x$ is rational.
Therefore you can define a surjection from the collection of all roots of rational numbers (which also includes all the rationals). So now comes the question, can you prove that this set is countable?
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
answered Dec 14 '18 at 4:46
Asaf Karagila♦Asaf Karagila
302k32427757
302k32427757
add a comment |
add a comment |
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