Construct bijections $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ and $f_2 :...












0












$begingroup$


I need to construct two bijections:
$$f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$$
$$f_2 : (mathbb{Z}times[0,1))rightarrowmathbb{R}$$
I know what bijection means and all conditions that functions have to fulfill in order to be bijective, but I have no idea how should I 'construct' them. I thought of drawing graphs of each set from function $f_1$, but it does not help me to do further steps.

It would be nice if you could show me step-by-step how it should be done.






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 15:57










  • $begingroup$
    I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 15:59










  • $begingroup$
    Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:06










  • $begingroup$
    I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 16:23










  • $begingroup$
    Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:33
















0












$begingroup$


I need to construct two bijections:
$$f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$$
$$f_2 : (mathbb{Z}times[0,1))rightarrowmathbb{R}$$
I know what bijection means and all conditions that functions have to fulfill in order to be bijective, but I have no idea how should I 'construct' them. I thought of drawing graphs of each set from function $f_1$, but it does not help me to do further steps.

It would be nice if you could show me step-by-step how it should be done.






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 15:57










  • $begingroup$
    I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 15:59










  • $begingroup$
    Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:06










  • $begingroup$
    I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 16:23










  • $begingroup$
    Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:33














0












0








0





$begingroup$


I need to construct two bijections:
$$f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$$
$$f_2 : (mathbb{Z}times[0,1))rightarrowmathbb{R}$$
I know what bijection means and all conditions that functions have to fulfill in order to be bijective, but I have no idea how should I 'construct' them. I thought of drawing graphs of each set from function $f_1$, but it does not help me to do further steps.

It would be nice if you could show me step-by-step how it should be done.






elementary-set-theory







share|cite|improve this question











$endgroup$




I need to construct two bijections:
$$f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$$
$$f_2 : (mathbb{Z}times[0,1))rightarrowmathbb{R}$$
I know what bijection means and all conditions that functions have to fulfill in order to be bijective, but I have no idea how should I 'construct' them. I thought of drawing graphs of each set from function $f_1$, but it does not help me to do further steps.

It would be nice if you could show me step-by-step how it should be done.






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 18:24







whiskeyo

















asked Dec 8 '18 at 15:48









whiskeyowhiskeyo

1107




1107












  • $begingroup$
    Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 15:57










  • $begingroup$
    I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 15:59










  • $begingroup$
    Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:06










  • $begingroup$
    I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 16:23










  • $begingroup$
    Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:33


















  • $begingroup$
    Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 15:57










  • $begingroup$
    I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 15:59










  • $begingroup$
    Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:06










  • $begingroup$
    I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
    $endgroup$
    – whiskeyo
    Dec 8 '18 at 16:23










  • $begingroup$
    Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
    $endgroup$
    – Henning Makholm
    Dec 8 '18 at 16:33
















$begingroup$
Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
$endgroup$
– Henning Makholm
Dec 8 '18 at 15:57




$begingroup$
Don't get too hung up on the word "construct". If you were just asked to find the bijections, would you know what to do?
$endgroup$
– Henning Makholm
Dec 8 '18 at 15:57












$begingroup$
I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
$endgroup$
– whiskeyo
Dec 8 '18 at 15:59




$begingroup$
I know that I should somehow find functions which fit these sets, but how can I do it if not by guessing?
$endgroup$
– whiskeyo
Dec 8 '18 at 15:59












$begingroup$
Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
$endgroup$
– Henning Makholm
Dec 8 '18 at 16:06




$begingroup$
Have you tried just "guessing"? This is not a follow-a- method exercise, it's just a check question to make sure you have understood what a bijection is.
$endgroup$
– Henning Makholm
Dec 8 '18 at 16:06












$begingroup$
I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
$endgroup$
– whiskeyo
Dec 8 '18 at 16:23




$begingroup$
I tried to transform $f_1 : ((0,1)times(2,3)) rightarrow (0,2)times(5,6)$ into $f_1 : A rightarrow B$, where $A : (0,1)rightarrow (2,3)$ and $B : (0,2)rightarrow (5,6)$ and then find functions fitting both sets, but I could not do that.
$endgroup$
– whiskeyo
Dec 8 '18 at 16:23












$begingroup$
Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
$endgroup$
– Henning Makholm
Dec 8 '18 at 16:33




$begingroup$
Hmmm. Can you make just a function that maps (0,1) bijectively to (2,3)?
$endgroup$
– Henning Makholm
Dec 8 '18 at 16:33










1 Answer
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$begingroup$

For $f_2$,The usual euclidean product of sets may be a little confusing here.



Try thinking of "$mathbb{Z}times[0,1)$" as the set "To each integer, assign an interval from 0 to 1." Then $f_2(x,t) = x+t$ is a bijection to $mathbb{R}$, where $xinmathbb{Z}$ is the integer part of some real number, and $tin[0,1)$ is the decimal part of that number.






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    $begingroup$

    For $f_2$,The usual euclidean product of sets may be a little confusing here.



    Try thinking of "$mathbb{Z}times[0,1)$" as the set "To each integer, assign an interval from 0 to 1." Then $f_2(x,t) = x+t$ is a bijection to $mathbb{R}$, where $xinmathbb{Z}$ is the integer part of some real number, and $tin[0,1)$ is the decimal part of that number.






    share|cite|improve this answer









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      1












      $begingroup$

      For $f_2$,The usual euclidean product of sets may be a little confusing here.



      Try thinking of "$mathbb{Z}times[0,1)$" as the set "To each integer, assign an interval from 0 to 1." Then $f_2(x,t) = x+t$ is a bijection to $mathbb{R}$, where $xinmathbb{Z}$ is the integer part of some real number, and $tin[0,1)$ is the decimal part of that number.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        For $f_2$,The usual euclidean product of sets may be a little confusing here.



        Try thinking of "$mathbb{Z}times[0,1)$" as the set "To each integer, assign an interval from 0 to 1." Then $f_2(x,t) = x+t$ is a bijection to $mathbb{R}$, where $xinmathbb{Z}$ is the integer part of some real number, and $tin[0,1)$ is the decimal part of that number.






        share|cite|improve this answer









        $endgroup$



        For $f_2$,The usual euclidean product of sets may be a little confusing here.



        Try thinking of "$mathbb{Z}times[0,1)$" as the set "To each integer, assign an interval from 0 to 1." Then $f_2(x,t) = x+t$ is a bijection to $mathbb{R}$, where $xinmathbb{Z}$ is the integer part of some real number, and $tin[0,1)$ is the decimal part of that number.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 18:37









        Adam CartisanoAdam Cartisano

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