Countable set - $aleph_0$ or finite?












1












$begingroup$


I have 100 sets $A_1,ldots,A_{100}$. They are all subsets of $Bbb R$. For each $A_i$ the complement of $A_i$ in $Bbb R$ is a countable set.



$A= A_1 cap A_2 cap ldots cap A_{100}$.



$B$ is the complement of $A$. What is the cardinal number of $B$?
There are 4 options as an answer:




  1. $0$

  2. finite number but not $0$

  3. $aleph_0$

  4. c


I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $Bbb R$ is a countable set' is all finite or one of them is $aleph_0$.



Am I right?



The formal answer for this question is $3$.






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
    $endgroup$
    – David K
    Jan 22 at 13:39










  • $begingroup$
    What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
    $endgroup$
    – saulspatz
    Jan 22 at 13:44










  • $begingroup$
    I mean ∩Ai , sorry
    $endgroup$
    – user3523226
    Jan 22 at 13:50










  • $begingroup$
    The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
    $endgroup$
    – Todor Markov
    Jan 22 at 13:55






  • 1




    $begingroup$
    If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
    $endgroup$
    – JuliusL33t
    Jan 22 at 13:55


















1












$begingroup$


I have 100 sets $A_1,ldots,A_{100}$. They are all subsets of $Bbb R$. For each $A_i$ the complement of $A_i$ in $Bbb R$ is a countable set.



$A= A_1 cap A_2 cap ldots cap A_{100}$.



$B$ is the complement of $A$. What is the cardinal number of $B$?
There are 4 options as an answer:




  1. $0$

  2. finite number but not $0$

  3. $aleph_0$

  4. c


I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $Bbb R$ is a countable set' is all finite or one of them is $aleph_0$.



Am I right?



The formal answer for this question is $3$.






discrete-mathematics







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
    $endgroup$
    – David K
    Jan 22 at 13:39










  • $begingroup$
    What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
    $endgroup$
    – saulspatz
    Jan 22 at 13:44










  • $begingroup$
    I mean ∩Ai , sorry
    $endgroup$
    – user3523226
    Jan 22 at 13:50










  • $begingroup$
    The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
    $endgroup$
    – Todor Markov
    Jan 22 at 13:55






  • 1




    $begingroup$
    If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
    $endgroup$
    – JuliusL33t
    Jan 22 at 13:55
















1












1








1





$begingroup$


I have 100 sets $A_1,ldots,A_{100}$. They are all subsets of $Bbb R$. For each $A_i$ the complement of $A_i$ in $Bbb R$ is a countable set.



$A= A_1 cap A_2 cap ldots cap A_{100}$.



$B$ is the complement of $A$. What is the cardinal number of $B$?
There are 4 options as an answer:




  1. $0$

  2. finite number but not $0$

  3. $aleph_0$

  4. c


I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $Bbb R$ is a countable set' is all finite or one of them is $aleph_0$.



Am I right?



The formal answer for this question is $3$.






discrete-mathematics







share|cite|improve this question











$endgroup$




I have 100 sets $A_1,ldots,A_{100}$. They are all subsets of $Bbb R$. For each $A_i$ the complement of $A_i$ in $Bbb R$ is a countable set.



$A= A_1 cap A_2 cap ldots cap A_{100}$.



$B$ is the complement of $A$. What is the cardinal number of $B$?
There are 4 options as an answer:




  1. $0$

  2. finite number but not $0$

  3. $aleph_0$

  4. c


I think that is can be $2$ or $3$ because we don't know if the 'complement of $A_i$ in $Bbb R$ is a countable set' is all finite or one of them is $aleph_0$.



Am I right?



The formal answer for this question is $3$.






discrete-mathematics




discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 13:49









TonyK

43k356135




43k356135










asked Jan 22 at 13:28









user3523226user3523226

647




647








  • 2




    $begingroup$
    I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
    $endgroup$
    – David K
    Jan 22 at 13:39










  • $begingroup$
    What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
    $endgroup$
    – saulspatz
    Jan 22 at 13:44










  • $begingroup$
    I mean ∩Ai , sorry
    $endgroup$
    – user3523226
    Jan 22 at 13:50










  • $begingroup$
    The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
    $endgroup$
    – Todor Markov
    Jan 22 at 13:55






  • 1




    $begingroup$
    If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
    $endgroup$
    – JuliusL33t
    Jan 22 at 13:55
















  • 2




    $begingroup$
    I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
    $endgroup$
    – David K
    Jan 22 at 13:39










  • $begingroup$
    What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
    $endgroup$
    – saulspatz
    Jan 22 at 13:44










  • $begingroup$
    I mean ∩Ai , sorry
    $endgroup$
    – user3523226
    Jan 22 at 13:50










  • $begingroup$
    The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
    $endgroup$
    – Todor Markov
    Jan 22 at 13:55






  • 1




    $begingroup$
    If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
    $endgroup$
    – JuliusL33t
    Jan 22 at 13:55










2




2




$begingroup$
I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
$endgroup$
– David K
Jan 22 at 13:39




$begingroup$
I think you're right. It would be easier for people to understand and answer the question if it the formulas were typeset with MathJax, however. See math.stackexchange.com/help/notation -- someone has already started to do this.
$endgroup$
– David K
Jan 22 at 13:39












$begingroup$
What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
$endgroup$
– saulspatz
Jan 22 at 13:44




$begingroup$
What does A=A1 ^ A2 ^ ... ^ A100 mean? Is this supposed to be a tower of exponents, or something else?
$endgroup$
– saulspatz
Jan 22 at 13:44












$begingroup$
I mean ∩Ai , sorry
$endgroup$
– user3523226
Jan 22 at 13:50




$begingroup$
I mean ∩Ai , sorry
$endgroup$
– user3523226
Jan 22 at 13:50












$begingroup$
The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
$endgroup$
– Todor Markov
Jan 22 at 13:55




$begingroup$
The complement of the intersection is the same as the union of the complements, so you have the union of 100 countable sets. Typically, this could be 0, finite, or countable infinite. However, in some literature countable is used to refers only to infinite sets, in which case it's 3.
$endgroup$
– Todor Markov
Jan 22 at 13:55




1




1




$begingroup$
If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
$endgroup$
– JuliusL33t
Jan 22 at 13:55






$begingroup$
If you take $A_i = mathbb{R}$ for all $i$, then $A = mathbb{R} $, each complement is empty (so countable) , and $B$ is empty with cardinality $0$.
$endgroup$
– JuliusL33t
Jan 22 at 13:55












2 Answers
2






active

oldest

votes


















1












$begingroup$

The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."



Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    By that reasoning can't $0$ also be an answer?
    $endgroup$
    – Keen-ameteur
    Jan 22 at 13:55










  • $begingroup$
    @Keen-ameteur: Yes! I have edited my answer accordingly.
    $endgroup$
    – TonyK
    Jan 22 at 14:20





















0












$begingroup$

If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since



$$B = A^c = (bigcap(A_i))^c = bigcup(A_i)^c$$



which is a finite union of exactly countable sets.
Otherwise, both $1$ and $2$ could be answers.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."



    Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      By that reasoning can't $0$ also be an answer?
      $endgroup$
      – Keen-ameteur
      Jan 22 at 13:55










    • $begingroup$
      @Keen-ameteur: Yes! I have edited my answer accordingly.
      $endgroup$
      – TonyK
      Jan 22 at 14:20


















    1












    $begingroup$

    The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."



    Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      By that reasoning can't $0$ also be an answer?
      $endgroup$
      – Keen-ameteur
      Jan 22 at 13:55










    • $begingroup$
      @Keen-ameteur: Yes! I have edited my answer accordingly.
      $endgroup$
      – TonyK
      Jan 22 at 14:20
















    1












    1








    1





    $begingroup$

    The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."



    Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.






    share|cite|improve this answer











    $endgroup$



    The answer depends only on what your definition of "countable" is. As you point out, if a countable set can be finite, then option $2$ is indeed a possibility. See this Wikipedia article: "Some authors use countable set to mean countably infinite alone."



    Edited to add: As Keen-amateur points out, option $1$ is also possible if a countable set can be empty.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 22 at 14:19

























    answered Jan 22 at 13:53









    TonyKTonyK

    43k356135




    43k356135












    • $begingroup$
      By that reasoning can't $0$ also be an answer?
      $endgroup$
      – Keen-ameteur
      Jan 22 at 13:55










    • $begingroup$
      @Keen-ameteur: Yes! I have edited my answer accordingly.
      $endgroup$
      – TonyK
      Jan 22 at 14:20




















    • $begingroup$
      By that reasoning can't $0$ also be an answer?
      $endgroup$
      – Keen-ameteur
      Jan 22 at 13:55










    • $begingroup$
      @Keen-ameteur: Yes! I have edited my answer accordingly.
      $endgroup$
      – TonyK
      Jan 22 at 14:20


















    $begingroup$
    By that reasoning can't $0$ also be an answer?
    $endgroup$
    – Keen-ameteur
    Jan 22 at 13:55




    $begingroup$
    By that reasoning can't $0$ also be an answer?
    $endgroup$
    – Keen-ameteur
    Jan 22 at 13:55












    $begingroup$
    @Keen-ameteur: Yes! I have edited my answer accordingly.
    $endgroup$
    – TonyK
    Jan 22 at 14:20






    $begingroup$
    @Keen-ameteur: Yes! I have edited my answer accordingly.
    $endgroup$
    – TonyK
    Jan 22 at 14:20













    0












    $begingroup$

    If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since



    $$B = A^c = (bigcap(A_i))^c = bigcup(A_i)^c$$



    which is a finite union of exactly countable sets.
    Otherwise, both $1$ and $2$ could be answers.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since



      $$B = A^c = (bigcap(A_i))^c = bigcup(A_i)^c$$



      which is a finite union of exactly countable sets.
      Otherwise, both $1$ and $2$ could be answers.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since



        $$B = A^c = (bigcap(A_i))^c = bigcup(A_i)^c$$



        which is a finite union of exactly countable sets.
        Otherwise, both $1$ and $2$ could be answers.






        share|cite|improve this answer









        $endgroup$



        If you take countable to mean exactly countable, i.e. not finite, and not empty then $3$, is your only answer since



        $$B = A^c = (bigcap(A_i))^c = bigcup(A_i)^c$$



        which is a finite union of exactly countable sets.
        Otherwise, both $1$ and $2$ could be answers.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 14:02









        JuliusL33tJuliusL33t

        1,3481917




        1,3481917






























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