Proving a set to be countable
up vote
3
down vote
favorite
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 1 hour ago
Aniruddha Deshmukh
806418
add a comment |
up vote
3
down vote
favorite
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 1 hour ago
Aniruddha Deshmukh
806418
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
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 1 hour ago
Aniruddha Deshmukh
806418
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 1 hour ago
Aniruddha Deshmukh
806418
asked 1 hour ago
Aniruddha Deshmukh
806418
asked 1 hour ago
Aniruddha Deshmukh
806418
asked 1 hour ago
Aniruddha Deshmukh
806418
asked 1 hour ago
Aniruddha Deshmukh
806418
806418
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
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
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
$$
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.
answered 1 hour ago
Martin Argerami
123k1176174
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
|
show 1 more comment
up vote
2
down vote
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 1 hour ago
Asaf Karagila♦
301k32422752
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
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
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
$$
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.
answered 1 hour ago
Martin Argerami
123k1176174
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
|
show 1 more comment
up vote
2
down vote
accepted
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
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
$$
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.
answered 1 hour ago
Martin Argerami
123k1176174
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
|
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
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
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
$$
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.
answered 1 hour ago
Martin Argerami
123k1176174
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
$$
S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
$$
$$
S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
$$
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.
answered 1 hour ago
Martin Argerami
123k1176174
edited 1 hour ago
answered 1 hour ago
Martin Argerami
123k1176174
answered 1 hour ago
Martin Argerami
123k1176174
answered 1 hour ago
Martin Argerami
123k1176174
123k1176174
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
|
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
Yes, you are right. I needed to duplicate the sets. It's done now.
– Martin Argerami
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
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
1 hour ago
|
show 1 more comment
up vote
2
down vote
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 1 hour ago
Asaf Karagila♦
301k32422752
add a comment |
up vote
2
down vote
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 1 hour ago
Asaf Karagila♦
301k32422752
add a comment |
up vote
2
down vote
up vote
2
down vote
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 1 hour ago
Asaf Karagila♦
301k32422752
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 1 hour ago
Asaf Karagila♦
301k32422752
answered 1 hour ago
Asaf Karagila♦
301k32422752
answered 1 hour ago
Asaf Karagila♦
301k32422752
answered 1 hour ago
Asaf Karagila♦
301k32422752
301k32422752
add a comment |
add a comment |
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