Proving a set to be countable











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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







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    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






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      asked 2 hours ago









      Aniruddha Deshmukh

      806418




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          2 Answers
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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
          $$
          S_1=bigcup_{ninmathbb N}bigcup_{xin E_n}left{(x,sqrt{tfrac1{n^2}-x^2})right}cupleft{(x,-sqrt{tfrac1{n^2}-x^2})right}.
          $$
          $$
          S_2=bigcup_{ninmathbb N}bigcup_{yin E_n}left{(sqrt{tfrac1{n^2}-y^2},y)right}cupleft{(-sqrt{tfrac1{n^2}-y^2}.y)right}.
          $$
          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.






          share|cite|improve this answer























          • 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
            2 hours 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
            2 hours ago












          • Yes, you are right. I needed to duplicate the sets. It's done now.
            – Martin Argerami
            2 hours 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
            2 hours 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
            2 hours ago


















          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?






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            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.






            share|cite|improve this answer























            • 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
              2 hours 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
              2 hours ago












            • Yes, you are right. I needed to duplicate the sets. It's done now.
              – Martin Argerami
              2 hours 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
              2 hours 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
              2 hours ago















            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.






            share|cite|improve this answer























            • 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
              2 hours 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
              2 hours ago












            • Yes, you are right. I needed to duplicate the sets. It's done now.
              – Martin Argerami
              2 hours 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
              2 hours 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
              2 hours ago













            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.






            share|cite|improve this answer














            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.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours 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
              2 hours 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
              2 hours ago












            • Yes, you are right. I needed to duplicate the sets. It's done now.
              – Martin Argerami
              2 hours 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
              2 hours 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
              2 hours 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
              2 hours 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
              2 hours ago












            • Yes, you are right. I needed to duplicate the sets. It's done now.
              – Martin Argerami
              2 hours 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
              2 hours 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
              2 hours 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
            2 hours 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
            2 hours 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
            2 hours 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
            2 hours ago














            Yes, you are right. I needed to duplicate the sets. It's done now.
            – Martin Argerami
            2 hours ago




            Yes, you are right. I needed to duplicate the sets. It's done now.
            – Martin Argerami
            2 hours 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
            2 hours 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
            2 hours 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
            2 hours 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
            2 hours ago










            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?






            share|cite|improve this answer

























              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?






              share|cite|improve this answer























                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?






                share|cite|improve this answer












                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?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                Asaf Karagila♦

                301k32422752




                301k32422752






























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