Show that the following pairs of sets have the same cardinality.












1















(a) Integers divisible by 3, and the even positive integers.



(b) $Bbb {R}$, and the interval $(0,infty)$.



(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).



(d) The intervals $(-infty, -1)$ and $(-1, 0)$.




I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.



Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.



Thank you.






real-analysis functions elementary-set-theory







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  • ok thank you, fixed it
    – smith
    Nov 27 '18 at 2:47










  • Do you mean the set $[5,6)cup [7,8)$?
    – TonyK
    Nov 27 '18 at 2:50










  • yes, it has been fixed
    – smith
    Nov 27 '18 at 2:51
















1















(a) Integers divisible by 3, and the even positive integers.



(b) $Bbb {R}$, and the interval $(0,infty)$.



(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).



(d) The intervals $(-infty, -1)$ and $(-1, 0)$.




I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.



Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.



Thank you.






real-analysis functions elementary-set-theory







share|cite|improve this question
























  • ok thank you, fixed it
    – smith
    Nov 27 '18 at 2:47










  • Do you mean the set $[5,6)cup [7,8)$?
    – TonyK
    Nov 27 '18 at 2:50










  • yes, it has been fixed
    – smith
    Nov 27 '18 at 2:51














1












1








1








(a) Integers divisible by 3, and the even positive integers.



(b) $Bbb {R}$, and the interval $(0,infty)$.



(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).



(d) The intervals $(-infty, -1)$ and $(-1, 0)$.




I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.



Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.



Thank you.






real-analysis functions elementary-set-theory







share|cite|improve this question
















(a) Integers divisible by 3, and the even positive integers.



(b) $Bbb {R}$, and the interval $(0,infty)$.



(c) The interval $[0, 2)$, and the set [5,6)∪[7,8).



(d) The intervals $(-infty, -1)$ and $(-1, 0)$.




I know that cardinality means that there is a bijection between the two sets, and that means there is a surjection and injection. For the first one I think you can simply make a function that is a bijection and prove that it is an injection and surjection. I am not sure how to do the last three.



Sorry for the poor formatting in advance and if you need more information about the question I am more than happy to provide it.



Thank you.






real-analysis functions elementary-set-theory




real-analysis functions elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 '18 at 21:28









greedoid

38.2k114797




38.2k114797










asked Nov 27 '18 at 2:44









smith

93




93












  • ok thank you, fixed it
    – smith
    Nov 27 '18 at 2:47










  • Do you mean the set $[5,6)cup [7,8)$?
    – TonyK
    Nov 27 '18 at 2:50










  • yes, it has been fixed
    – smith
    Nov 27 '18 at 2:51


















  • ok thank you, fixed it
    – smith
    Nov 27 '18 at 2:47










  • Do you mean the set $[5,6)cup [7,8)$?
    – TonyK
    Nov 27 '18 at 2:50










  • yes, it has been fixed
    – smith
    Nov 27 '18 at 2:51
















ok thank you, fixed it
– smith
Nov 27 '18 at 2:47




ok thank you, fixed it
– smith
Nov 27 '18 at 2:47












Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50




Do you mean the set $[5,6)cup [7,8)$?
– TonyK
Nov 27 '18 at 2:50












yes, it has been fixed
– smith
Nov 27 '18 at 2:51




yes, it has been fixed
– smith
Nov 27 '18 at 2:51










4 Answers
4






active

oldest

votes


















1














Hints:



(b) Map $x in mathbb{R}$ by $x to e^x$



(c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$



(d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$



Show that these are all bijections.






share|cite|improve this answer





























    0














    (b) $mathbb{R}$, and the interval (0,∞).



    Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$



    (c) The interval [0,2), and the set [5,6) or [7,8).



    Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.



    (d) The intervals (−∞,−1) and (−1,0).



    Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$






    share|cite|improve this answer





























      0














      a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$



      and then $g:mathbb{Z}to C$ defined by:



      $$
      g(n)
      begin{cases}
      = 4n+2 & text{if } ngeq 0, \
      = 4|n| & text{if } n < 0. \
      end{cases}
      $$



      Then let $gcirc f$ will do the job.






      share|cite|improve this answer































        0














        The bijections, reading left to right, can be chosen as:



        (a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.






        share|cite|improve this answer





















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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Hints:



          (b) Map $x in mathbb{R}$ by $x to e^x$



          (c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$



          (d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$



          Show that these are all bijections.






          share|cite|improve this answer


























            1














            Hints:



            (b) Map $x in mathbb{R}$ by $x to e^x$



            (c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$



            (d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$



            Show that these are all bijections.






            share|cite|improve this answer
























              1












              1








              1






              Hints:



              (b) Map $x in mathbb{R}$ by $x to e^x$



              (c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$



              (d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$



              Show that these are all bijections.






              share|cite|improve this answer












              Hints:



              (b) Map $x in mathbb{R}$ by $x to e^x$



              (c) Map $x in [0, 2)$ by $x to x + 5$ if $x in [0,1)$ and $x to x + 6$ if $x in [1,2)$



              (d) Map $x in (-infty,-1)$ by $x to frac{1}{x}$



              Show that these are all bijections.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 27 '18 at 3:03









              AlkaKadri

              1,459411




              1,459411























                  0














                  (b) $mathbb{R}$, and the interval (0,∞).



                  Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$



                  (c) The interval [0,2), and the set [5,6) or [7,8).



                  Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.



                  (d) The intervals (−∞,−1) and (−1,0).



                  Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$






                  share|cite|improve this answer


























                    0














                    (b) $mathbb{R}$, and the interval (0,∞).



                    Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$



                    (c) The interval [0,2), and the set [5,6) or [7,8).



                    Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.



                    (d) The intervals (−∞,−1) and (−1,0).



                    Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$






                    share|cite|improve this answer
























                      0












                      0








                      0






                      (b) $mathbb{R}$, and the interval (0,∞).



                      Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$



                      (c) The interval [0,2), and the set [5,6) or [7,8).



                      Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.



                      (d) The intervals (−∞,−1) and (−1,0).



                      Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$






                      share|cite|improve this answer












                      (b) $mathbb{R}$, and the interval (0,∞).



                      Consider $f : mathbb{R}rightarrow (0,infty)$ defined by $f(x)=e^x.$



                      (c) The interval [0,2), and the set [5,6) or [7,8).



                      Consider $g: [0,2)rightarrow [5,6)$ defined by $g(x)=5+frac{x}{2}$ or $g'(x)=7+frac{x}{2}$ if codomain is $[7,8)$.



                      (d) The intervals (−∞,−1) and (−1,0).



                      Consider $h: (-1,0)rightarrow (-infty,-1)$ defined by $h(x)=tan(frac{pi}{2}x)-1.$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Nov 27 '18 at 3:02









                      LeB

                      986217




                      986217























                          0














                          a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$



                          and then $g:mathbb{Z}to C$ defined by:



                          $$
                          g(n)
                          begin{cases}
                          = 4n+2 & text{if } ngeq 0, \
                          = 4|n| & text{if } n < 0. \
                          end{cases}
                          $$



                          Then let $gcirc f$ will do the job.






                          share|cite|improve this answer




























                            0














                            a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$



                            and then $g:mathbb{Z}to C$ defined by:



                            $$
                            g(n)
                            begin{cases}
                            = 4n+2 & text{if } ngeq 0, \
                            = 4|n| & text{if } n < 0. \
                            end{cases}
                            $$



                            Then let $gcirc f$ will do the job.






                            share|cite|improve this answer


























                              0












                              0








                              0






                              a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$



                              and then $g:mathbb{Z}to C$ defined by:



                              $$
                              g(n)
                              begin{cases}
                              = 4n+2 & text{if } ngeq 0, \
                              = 4|n| & text{if } n < 0. \
                              end{cases}
                              $$



                              Then let $gcirc f$ will do the job.






                              share|cite|improve this answer














                              a) Let $A$ be a set of integers divisible by 3 and $C$ a set of positive even integers. You can take a composition of following functions. First $f:Ato mathbb{Z}$ defined with $$nlongmapsto {nover 3}$$



                              and then $g:mathbb{Z}to C$ defined by:



                              $$
                              g(n)
                              begin{cases}
                              = 4n+2 & text{if } ngeq 0, \
                              = 4|n| & text{if } n < 0. \
                              end{cases}
                              $$



                              Then let $gcirc f$ will do the job.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Nov 27 '18 at 22:03

























                              answered Nov 27 '18 at 21:57









                              greedoid

                              38.2k114797




                              38.2k114797























                                  0














                                  The bijections, reading left to right, can be chosen as:



                                  (a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.






                                  share|cite|improve this answer


























                                    0














                                    The bijections, reading left to right, can be chosen as:



                                    (a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.






                                    share|cite|improve this answer
























                                      0












                                      0








                                      0






                                      The bijections, reading left to right, can be chosen as:



                                      (a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.






                                      share|cite|improve this answer












                                      The bijections, reading left to right, can be chosen as:



                                      (a) $|4x/3|+2[xge 0]$ (b) $x+sqrt{1+x^2}$ (c) $x+5+lfloor xrfloor$ (d) $1/x$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Nov 27 '18 at 22:11









                                      J.G.

                                      23.2k22137




                                      23.2k22137






























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