Suppose that $A$ and $B$ are sets and that $f: A rightarrow B$ is onto. Does being onto guarantee the sets...












1












$begingroup$



Suppose that $A$ and $B$ are sets and that $f: A rightarrow B$ is onto. Determine which of the following statements are true:




  • If $A$ is finite then $B$ is finite.

  • If $B$ is finite, then $A$ is finite.




My defintion for onto is: a function $f: X rightarrow Y$ is said to be onto provided for each $ y in Y$ there exists at least one $x in X$ such that $f(x)= y$. I also have the statement "Thus a function is onto if the range is equal to the codomain."



I am thinking that if $B$ is finite then $A$ is finite is the true statement and if $A$ is finite then $B$ is finite is the false statement. I think this statement is false.



Is this correct? How do I begin a proof or give a counterexample?






functions elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    "There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
    $endgroup$
    – Jason DeVito
    May 12 '15 at 0:39










  • $begingroup$
    In fact, the first statement is true, whereas the second is false.
    $endgroup$
    – Omnomnomnom
    May 12 '15 at 0:41






  • 1




    $begingroup$
    You need to stop thinking about this problem for a little while and go learn what a function is.
    $endgroup$
    – WillO
    May 12 '15 at 1:18






  • 1




    $begingroup$
    @WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
    $endgroup$
    – user219081
    May 12 '15 at 1:22










  • $begingroup$
    @AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
    $endgroup$
    – WillO
    May 12 '15 at 2:51
















1












$begingroup$



Suppose that $A$ and $B$ are sets and that $f: A rightarrow B$ is onto. Determine which of the following statements are true:




  • If $A$ is finite then $B$ is finite.

  • If $B$ is finite, then $A$ is finite.




My defintion for onto is: a function $f: X rightarrow Y$ is said to be onto provided for each $ y in Y$ there exists at least one $x in X$ such that $f(x)= y$. I also have the statement "Thus a function is onto if the range is equal to the codomain."



I am thinking that if $B$ is finite then $A$ is finite is the true statement and if $A$ is finite then $B$ is finite is the false statement. I think this statement is false.



Is this correct? How do I begin a proof or give a counterexample?






functions elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    "There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
    $endgroup$
    – Jason DeVito
    May 12 '15 at 0:39










  • $begingroup$
    In fact, the first statement is true, whereas the second is false.
    $endgroup$
    – Omnomnomnom
    May 12 '15 at 0:41






  • 1




    $begingroup$
    You need to stop thinking about this problem for a little while and go learn what a function is.
    $endgroup$
    – WillO
    May 12 '15 at 1:18






  • 1




    $begingroup$
    @WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
    $endgroup$
    – user219081
    May 12 '15 at 1:22










  • $begingroup$
    @AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
    $endgroup$
    – WillO
    May 12 '15 at 2:51














1












1








1





$begingroup$



Suppose that $A$ and $B$ are sets and that $f: A rightarrow B$ is onto. Determine which of the following statements are true:




  • If $A$ is finite then $B$ is finite.

  • If $B$ is finite, then $A$ is finite.




My defintion for onto is: a function $f: X rightarrow Y$ is said to be onto provided for each $ y in Y$ there exists at least one $x in X$ such that $f(x)= y$. I also have the statement "Thus a function is onto if the range is equal to the codomain."



I am thinking that if $B$ is finite then $A$ is finite is the true statement and if $A$ is finite then $B$ is finite is the false statement. I think this statement is false.



Is this correct? How do I begin a proof or give a counterexample?






functions elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$





Suppose that $A$ and $B$ are sets and that $f: A rightarrow B$ is onto. Determine which of the following statements are true:




  • If $A$ is finite then $B$ is finite.

  • If $B$ is finite, then $A$ is finite.




My defintion for onto is: a function $f: X rightarrow Y$ is said to be onto provided for each $ y in Y$ there exists at least one $x in X$ such that $f(x)= y$. I also have the statement "Thus a function is onto if the range is equal to the codomain."



I am thinking that if $B$ is finite then $A$ is finite is the true statement and if $A$ is finite then $B$ is finite is the false statement. I think this statement is false.



Is this correct? How do I begin a proof or give a counterexample?






functions elementary-set-theory proof-writing




functions elementary-set-theory proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 12 '15 at 1:21

























asked May 12 '15 at 0:30







user219081















  • 2




    $begingroup$
    "There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
    $endgroup$
    – Jason DeVito
    May 12 '15 at 0:39










  • $begingroup$
    In fact, the first statement is true, whereas the second is false.
    $endgroup$
    – Omnomnomnom
    May 12 '15 at 0:41






  • 1




    $begingroup$
    You need to stop thinking about this problem for a little while and go learn what a function is.
    $endgroup$
    – WillO
    May 12 '15 at 1:18






  • 1




    $begingroup$
    @WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
    $endgroup$
    – user219081
    May 12 '15 at 1:22










  • $begingroup$
    @AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
    $endgroup$
    – WillO
    May 12 '15 at 2:51














  • 2




    $begingroup$
    "There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
    $endgroup$
    – Jason DeVito
    May 12 '15 at 0:39










  • $begingroup$
    In fact, the first statement is true, whereas the second is false.
    $endgroup$
    – Omnomnomnom
    May 12 '15 at 0:41






  • 1




    $begingroup$
    You need to stop thinking about this problem for a little while and go learn what a function is.
    $endgroup$
    – WillO
    May 12 '15 at 1:18






  • 1




    $begingroup$
    @WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
    $endgroup$
    – user219081
    May 12 '15 at 1:22










  • $begingroup$
    @AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
    $endgroup$
    – WillO
    May 12 '15 at 2:51








2




2




$begingroup$
"There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
$endgroup$
– Jason DeVito
May 12 '15 at 0:39




$begingroup$
"There is no guarantee that an onto function sends one $x$ value to one and only one $y$ value" - actually, the "function" part of "onto function" guarantees this. Why don't you write down some examples of $A$s and $B$s and refine your conjecture?
$endgroup$
– Jason DeVito
May 12 '15 at 0:39












$begingroup$
In fact, the first statement is true, whereas the second is false.
$endgroup$
– Omnomnomnom
May 12 '15 at 0:41




$begingroup$
In fact, the first statement is true, whereas the second is false.
$endgroup$
– Omnomnomnom
May 12 '15 at 0:41




1




1




$begingroup$
You need to stop thinking about this problem for a little while and go learn what a function is.
$endgroup$
– WillO
May 12 '15 at 1:18




$begingroup$
You need to stop thinking about this problem for a little while and go learn what a function is.
$endgroup$
– WillO
May 12 '15 at 1:18




1




1




$begingroup$
@WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
$endgroup$
– user219081
May 12 '15 at 1:22




$begingroup$
@WillO thank you for that great comment. I realized what I wrote didn't make sense after posting. Thanks.
$endgroup$
– user219081
May 12 '15 at 1:22












$begingroup$
@AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
$endgroup$
– WillO
May 12 '15 at 2:51




$begingroup$
@AlyssaWallace: Thank you for taking the comment seriously. Best of luck with all your studies.
$endgroup$
– WillO
May 12 '15 at 2:51










5 Answers
5






active

oldest

votes


















0












$begingroup$

An onto function need not be one-to-one. Figure out which side doesn't need to be the "one".






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Consider $A = Bbb N$, and consider $B = {0,1,2,dots,b-1}$, for some fixed positive integer $b$.



    You should know that for any natural number $a$, we can write:



    $a = qb + r$, where $r in B$, and that the natural numbers $q,r$ (also known as "quotient and remainder") are uniquely determined by $a$ and $b$.



    Thus we can define $f:A to B$ by:



    $f(a) = r$, and this function is onto. But $B$ is finite, and $A$ is infinite.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Or simpler yet, you could define $f(x)=min(x,b-1)$.
      $endgroup$
      – Henning Makholm
      May 12 '15 at 0:49










    • $begingroup$
      Indeed, there are many projections of an infinite set onto a finite one.
      $endgroup$
      – David Wheeler
      May 12 '15 at 3:53



















    0












    $begingroup$

    Ok, assume first that $A = { x_1, ldots, x_n }$ is finite. Since $f$ is onto, we have $f(A) = B$. From this it is easy to see that $B$ must be finite.



    On the other hand, if $B = { x_1, ldots, x_m }$ is finite, you could for example construct a function $f: mathbb N to B$, by setting $f(n) = x_n$ for $n = 1, ldots, m$, and $f(n) = x_1$ for $n in mathbb N backslash {1, ldots, m}$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you elaborate on "from this it is easy to see that B must be infinite?"
      $endgroup$
      – user219081
      May 12 '15 at 0:58










    • $begingroup$
      Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
      $endgroup$
      – aexl
      May 12 '15 at 1:02





















    0












    $begingroup$

    If $f:A rightarrow B$ is an onto map then $|B|leq |A|.$ So if $A$ is a finite set then $B$ is also a finite set.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can you please explain how you make the inequality statement?
      $endgroup$
      – user219081
      May 12 '15 at 0:57










    • $begingroup$
      Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
      $endgroup$
      – neelkanth
      May 12 '15 at 1:02










    • $begingroup$
      And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
      $endgroup$
      – neelkanth
      May 12 '15 at 1:03



















    0












    $begingroup$

    Statement 2 is not always necessarily true.



    Consider a function floor()%finitenumber which maps real numbers to integer set. Then A can be infinite but B is always finite.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      The first statement is necessarily always true, though.
      $endgroup$
      – Henning Makholm
      May 12 '15 at 0:53












    • $begingroup$
      @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
      $endgroup$
      – Henning Makholm
      May 12 '15 at 0:58













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    5 Answers
    5






    active

    oldest

    votes








    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    An onto function need not be one-to-one. Figure out which side doesn't need to be the "one".






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      An onto function need not be one-to-one. Figure out which side doesn't need to be the "one".






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        An onto function need not be one-to-one. Figure out which side doesn't need to be the "one".






        share|cite|improve this answer









        $endgroup$



        An onto function need not be one-to-one. Figure out which side doesn't need to be the "one".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 12 '15 at 0:41









        parallaxeffectparallaxeffect

        1165




        1165























            0












            $begingroup$

            Consider $A = Bbb N$, and consider $B = {0,1,2,dots,b-1}$, for some fixed positive integer $b$.



            You should know that for any natural number $a$, we can write:



            $a = qb + r$, where $r in B$, and that the natural numbers $q,r$ (also known as "quotient and remainder") are uniquely determined by $a$ and $b$.



            Thus we can define $f:A to B$ by:



            $f(a) = r$, and this function is onto. But $B$ is finite, and $A$ is infinite.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or simpler yet, you could define $f(x)=min(x,b-1)$.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:49










            • $begingroup$
              Indeed, there are many projections of an infinite set onto a finite one.
              $endgroup$
              – David Wheeler
              May 12 '15 at 3:53
















            0












            $begingroup$

            Consider $A = Bbb N$, and consider $B = {0,1,2,dots,b-1}$, for some fixed positive integer $b$.



            You should know that for any natural number $a$, we can write:



            $a = qb + r$, where $r in B$, and that the natural numbers $q,r$ (also known as "quotient and remainder") are uniquely determined by $a$ and $b$.



            Thus we can define $f:A to B$ by:



            $f(a) = r$, and this function is onto. But $B$ is finite, and $A$ is infinite.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Or simpler yet, you could define $f(x)=min(x,b-1)$.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:49










            • $begingroup$
              Indeed, there are many projections of an infinite set onto a finite one.
              $endgroup$
              – David Wheeler
              May 12 '15 at 3:53














            0












            0








            0





            $begingroup$

            Consider $A = Bbb N$, and consider $B = {0,1,2,dots,b-1}$, for some fixed positive integer $b$.



            You should know that for any natural number $a$, we can write:



            $a = qb + r$, where $r in B$, and that the natural numbers $q,r$ (also known as "quotient and remainder") are uniquely determined by $a$ and $b$.



            Thus we can define $f:A to B$ by:



            $f(a) = r$, and this function is onto. But $B$ is finite, and $A$ is infinite.






            share|cite|improve this answer









            $endgroup$



            Consider $A = Bbb N$, and consider $B = {0,1,2,dots,b-1}$, for some fixed positive integer $b$.



            You should know that for any natural number $a$, we can write:



            $a = qb + r$, where $r in B$, and that the natural numbers $q,r$ (also known as "quotient and remainder") are uniquely determined by $a$ and $b$.



            Thus we can define $f:A to B$ by:



            $f(a) = r$, and this function is onto. But $B$ is finite, and $A$ is infinite.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 12 '15 at 0:47









            David WheelerDavid Wheeler

            12.7k11730




            12.7k11730












            • $begingroup$
              Or simpler yet, you could define $f(x)=min(x,b-1)$.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:49










            • $begingroup$
              Indeed, there are many projections of an infinite set onto a finite one.
              $endgroup$
              – David Wheeler
              May 12 '15 at 3:53


















            • $begingroup$
              Or simpler yet, you could define $f(x)=min(x,b-1)$.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:49










            • $begingroup$
              Indeed, there are many projections of an infinite set onto a finite one.
              $endgroup$
              – David Wheeler
              May 12 '15 at 3:53
















            $begingroup$
            Or simpler yet, you could define $f(x)=min(x,b-1)$.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:49




            $begingroup$
            Or simpler yet, you could define $f(x)=min(x,b-1)$.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:49












            $begingroup$
            Indeed, there are many projections of an infinite set onto a finite one.
            $endgroup$
            – David Wheeler
            May 12 '15 at 3:53




            $begingroup$
            Indeed, there are many projections of an infinite set onto a finite one.
            $endgroup$
            – David Wheeler
            May 12 '15 at 3:53











            0












            $begingroup$

            Ok, assume first that $A = { x_1, ldots, x_n }$ is finite. Since $f$ is onto, we have $f(A) = B$. From this it is easy to see that $B$ must be finite.



            On the other hand, if $B = { x_1, ldots, x_m }$ is finite, you could for example construct a function $f: mathbb N to B$, by setting $f(n) = x_n$ for $n = 1, ldots, m$, and $f(n) = x_1$ for $n in mathbb N backslash {1, ldots, m}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you elaborate on "from this it is easy to see that B must be infinite?"
              $endgroup$
              – user219081
              May 12 '15 at 0:58










            • $begingroup$
              Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
              $endgroup$
              – aexl
              May 12 '15 at 1:02


















            0












            $begingroup$

            Ok, assume first that $A = { x_1, ldots, x_n }$ is finite. Since $f$ is onto, we have $f(A) = B$. From this it is easy to see that $B$ must be finite.



            On the other hand, if $B = { x_1, ldots, x_m }$ is finite, you could for example construct a function $f: mathbb N to B$, by setting $f(n) = x_n$ for $n = 1, ldots, m$, and $f(n) = x_1$ for $n in mathbb N backslash {1, ldots, m}$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you elaborate on "from this it is easy to see that B must be infinite?"
              $endgroup$
              – user219081
              May 12 '15 at 0:58










            • $begingroup$
              Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
              $endgroup$
              – aexl
              May 12 '15 at 1:02
















            0












            0








            0





            $begingroup$

            Ok, assume first that $A = { x_1, ldots, x_n }$ is finite. Since $f$ is onto, we have $f(A) = B$. From this it is easy to see that $B$ must be finite.



            On the other hand, if $B = { x_1, ldots, x_m }$ is finite, you could for example construct a function $f: mathbb N to B$, by setting $f(n) = x_n$ for $n = 1, ldots, m$, and $f(n) = x_1$ for $n in mathbb N backslash {1, ldots, m}$.






            share|cite|improve this answer









            $endgroup$



            Ok, assume first that $A = { x_1, ldots, x_n }$ is finite. Since $f$ is onto, we have $f(A) = B$. From this it is easy to see that $B$ must be finite.



            On the other hand, if $B = { x_1, ldots, x_m }$ is finite, you could for example construct a function $f: mathbb N to B$, by setting $f(n) = x_n$ for $n = 1, ldots, m$, and $f(n) = x_1$ for $n in mathbb N backslash {1, ldots, m}$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 12 '15 at 0:52









            aexlaexl

            1,287717




            1,287717












            • $begingroup$
              Can you elaborate on "from this it is easy to see that B must be infinite?"
              $endgroup$
              – user219081
              May 12 '15 at 0:58










            • $begingroup$
              Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
              $endgroup$
              – aexl
              May 12 '15 at 1:02




















            • $begingroup$
              Can you elaborate on "from this it is easy to see that B must be infinite?"
              $endgroup$
              – user219081
              May 12 '15 at 0:58










            • $begingroup$
              Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
              $endgroup$
              – aexl
              May 12 '15 at 1:02


















            $begingroup$
            Can you elaborate on "from this it is easy to see that B must be infinite?"
            $endgroup$
            – user219081
            May 12 '15 at 0:58




            $begingroup$
            Can you elaborate on "from this it is easy to see that B must be infinite?"
            $endgroup$
            – user219081
            May 12 '15 at 0:58












            $begingroup$
            Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
            $endgroup$
            – aexl
            May 12 '15 at 1:02






            $begingroup$
            Yeah, I mean we obviously have $f(A) = { f(x_1), ldots, f(x_n) } = B$, which means that $vert B vert leq n$.
            $endgroup$
            – aexl
            May 12 '15 at 1:02













            0












            $begingroup$

            If $f:A rightarrow B$ is an onto map then $|B|leq |A|.$ So if $A$ is a finite set then $B$ is also a finite set.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you please explain how you make the inequality statement?
              $endgroup$
              – user219081
              May 12 '15 at 0:57










            • $begingroup$
              Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:02










            • $begingroup$
              And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:03
















            0












            $begingroup$

            If $f:A rightarrow B$ is an onto map then $|B|leq |A|.$ So if $A$ is a finite set then $B$ is also a finite set.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Can you please explain how you make the inequality statement?
              $endgroup$
              – user219081
              May 12 '15 at 0:57










            • $begingroup$
              Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:02










            • $begingroup$
              And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:03














            0












            0








            0





            $begingroup$

            If $f:A rightarrow B$ is an onto map then $|B|leq |A|.$ So if $A$ is a finite set then $B$ is also a finite set.






            share|cite|improve this answer









            $endgroup$



            If $f:A rightarrow B$ is an onto map then $|B|leq |A|.$ So if $A$ is a finite set then $B$ is also a finite set.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 12 '15 at 0:54









            neelkanthneelkanth

            2,27621029




            2,27621029












            • $begingroup$
              Can you please explain how you make the inequality statement?
              $endgroup$
              – user219081
              May 12 '15 at 0:57










            • $begingroup$
              Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:02










            • $begingroup$
              And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:03


















            • $begingroup$
              Can you please explain how you make the inequality statement?
              $endgroup$
              – user219081
              May 12 '15 at 0:57










            • $begingroup$
              Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:02










            • $begingroup$
              And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
              $endgroup$
              – neelkanth
              May 12 '15 at 1:03
















            $begingroup$
            Can you please explain how you make the inequality statement?
            $endgroup$
            – user219081
            May 12 '15 at 0:57




            $begingroup$
            Can you please explain how you make the inequality statement?
            $endgroup$
            – user219081
            May 12 '15 at 0:57












            $begingroup$
            Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
            $endgroup$
            – neelkanth
            May 12 '15 at 1:02




            $begingroup$
            Actually the proof is by Axiom of Choice that if there as an onto map from A to B then there is one one map from B to A.
            $endgroup$
            – neelkanth
            May 12 '15 at 1:02












            $begingroup$
            And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
            $endgroup$
            – neelkanth
            May 12 '15 at 1:03




            $begingroup$
            And if there is one one map from B to A then B is equivalent to a subset of A so cardinality of B is less than or equal to that of A.
            $endgroup$
            – neelkanth
            May 12 '15 at 1:03











            0












            $begingroup$

            Statement 2 is not always necessarily true.



            Consider a function floor()%finitenumber which maps real numbers to integer set. Then A can be infinite but B is always finite.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The first statement is necessarily always true, though.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:53












            • $begingroup$
              @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:58


















            0












            $begingroup$

            Statement 2 is not always necessarily true.



            Consider a function floor()%finitenumber which maps real numbers to integer set. Then A can be infinite but B is always finite.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The first statement is necessarily always true, though.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:53












            • $begingroup$
              @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:58
















            0












            0








            0





            $begingroup$

            Statement 2 is not always necessarily true.



            Consider a function floor()%finitenumber which maps real numbers to integer set. Then A can be infinite but B is always finite.






            share|cite|improve this answer











            $endgroup$



            Statement 2 is not always necessarily true.



            Consider a function floor()%finitenumber which maps real numbers to integer set. Then A can be infinite but B is always finite.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 12 '15 at 1:15

























            answered May 12 '15 at 0:52









            SunnySunny

            84




            84












            • $begingroup$
              The first statement is necessarily always true, though.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:53












            • $begingroup$
              @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:58




















            • $begingroup$
              The first statement is necessarily always true, though.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:53












            • $begingroup$
              @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
              $endgroup$
              – Henning Makholm
              May 12 '15 at 0:58


















            $begingroup$
            The first statement is necessarily always true, though.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:53






            $begingroup$
            The first statement is necessarily always true, though.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:53














            $begingroup$
            @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:58






            $begingroup$
            @Ilham: Sorry, I mistyped. In any case, it looked as if you are claiming that each of the statements is sometimes-untrue.
            $endgroup$
            – Henning Makholm
            May 12 '15 at 0:58




















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