What Does the Following Notation Mean












1












$begingroup$


Currently working on some set theory problems, in particular I need to prove that



$|[0, 1]^mathbb{N}| = |{0, 1}^{mathbb{N} times mathbb{N}}|$



First let me say I'm not looking for a solution to the problem, as this is an assignment question, I only provide it as context for the level of understanding I'm at.



All I know, is that ${0, 1}^{mathbb{N} times mathbb{N}}$ is the set of all functions $F:mathbb{N} times mathbb{N} rightarrow {0, 1}$



I'm not sure what such a function would look like. My initial thought was that it maps a pair of natural numbers, to an ordered pair of the form (0, 1) but I realized pretty quickly that doesn't make much sense. Does it map a pair of natural numbers, to a pair of strings of 0s and 1s of any size?



If someone could give me an example of a function, and point me to a good resource for some of this notation, that'd be fantastic. My textbook seems to mostly take the notation for granted.






discrete-mathematics notation







share|cite|improve this question











$endgroup$












  • $begingroup$
    ${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 4:19






  • 1




    $begingroup$
    Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
    $endgroup$
    – Vortron
    Dec 6 '18 at 4:27
















1












$begingroup$


Currently working on some set theory problems, in particular I need to prove that



$|[0, 1]^mathbb{N}| = |{0, 1}^{mathbb{N} times mathbb{N}}|$



First let me say I'm not looking for a solution to the problem, as this is an assignment question, I only provide it as context for the level of understanding I'm at.



All I know, is that ${0, 1}^{mathbb{N} times mathbb{N}}$ is the set of all functions $F:mathbb{N} times mathbb{N} rightarrow {0, 1}$



I'm not sure what such a function would look like. My initial thought was that it maps a pair of natural numbers, to an ordered pair of the form (0, 1) but I realized pretty quickly that doesn't make much sense. Does it map a pair of natural numbers, to a pair of strings of 0s and 1s of any size?



If someone could give me an example of a function, and point me to a good resource for some of this notation, that'd be fantastic. My textbook seems to mostly take the notation for granted.






discrete-mathematics notation







share|cite|improve this question











$endgroup$












  • $begingroup$
    ${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 4:19






  • 1




    $begingroup$
    Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
    $endgroup$
    – Vortron
    Dec 6 '18 at 4:27














1












1








1





$begingroup$


Currently working on some set theory problems, in particular I need to prove that



$|[0, 1]^mathbb{N}| = |{0, 1}^{mathbb{N} times mathbb{N}}|$



First let me say I'm not looking for a solution to the problem, as this is an assignment question, I only provide it as context for the level of understanding I'm at.



All I know, is that ${0, 1}^{mathbb{N} times mathbb{N}}$ is the set of all functions $F:mathbb{N} times mathbb{N} rightarrow {0, 1}$



I'm not sure what such a function would look like. My initial thought was that it maps a pair of natural numbers, to an ordered pair of the form (0, 1) but I realized pretty quickly that doesn't make much sense. Does it map a pair of natural numbers, to a pair of strings of 0s and 1s of any size?



If someone could give me an example of a function, and point me to a good resource for some of this notation, that'd be fantastic. My textbook seems to mostly take the notation for granted.






discrete-mathematics notation







share|cite|improve this question











$endgroup$




Currently working on some set theory problems, in particular I need to prove that



$|[0, 1]^mathbb{N}| = |{0, 1}^{mathbb{N} times mathbb{N}}|$



First let me say I'm not looking for a solution to the problem, as this is an assignment question, I only provide it as context for the level of understanding I'm at.



All I know, is that ${0, 1}^{mathbb{N} times mathbb{N}}$ is the set of all functions $F:mathbb{N} times mathbb{N} rightarrow {0, 1}$



I'm not sure what such a function would look like. My initial thought was that it maps a pair of natural numbers, to an ordered pair of the form (0, 1) but I realized pretty quickly that doesn't make much sense. Does it map a pair of natural numbers, to a pair of strings of 0s and 1s of any size?



If someone could give me an example of a function, and point me to a good resource for some of this notation, that'd be fantastic. My textbook seems to mostly take the notation for granted.






discrete-mathematics notation




discrete-mathematics notation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 4:25









Andrés E. Caicedo

65.3k8158247




65.3k8158247










asked Dec 6 '18 at 4:13









VortronVortron

184




184












  • $begingroup$
    ${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 4:19






  • 1




    $begingroup$
    Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
    $endgroup$
    – Vortron
    Dec 6 '18 at 4:27


















  • $begingroup$
    ${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
    $endgroup$
    – Thomas Shelby
    Dec 6 '18 at 4:19






  • 1




    $begingroup$
    Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
    $endgroup$
    – Vortron
    Dec 6 '18 at 4:27
















$begingroup$
${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
$endgroup$
– Thomas Shelby
Dec 6 '18 at 4:19




$begingroup$
${0, 1}^{mathbb{N}}$ denotes the set of all sequences consisting only of $0$'s and $1$'s.
$endgroup$
– Thomas Shelby
Dec 6 '18 at 4:19




1




1




$begingroup$
Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
$endgroup$
– Vortron
Dec 6 '18 at 4:27




$begingroup$
Unfortunately not. I think, though, that your first comment answered my question. It seems my original interpretation was correct.
$endgroup$
– Vortron
Dec 6 '18 at 4:27










1 Answer
1






active

oldest

votes


















2












$begingroup$

A function $F:mathbb{N}times mathbb{N} to {0,1}$ (“from $mathbb{N}times mathbb{N}$ to ${0,1}$”) assigns to each element of the set $mathbb{N}times mathbb{N}$, ordered pairs of integers, an element in the set {0,1}, either 0 or 1. So if $m$ and $n$ are natural numbers, $F((m,n))=0$ or $F((m,n))=1$.



Some examples: let $F:mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 0$. On the left of “$to$”, we have the domain of our function. On the right, we have our codomain. On the left of “$mapsto$”, we have some arbitrary element in the domain. On the right, we have the explicit element in the codomain that $(m,n)$ will map to. So, in particular, $F((m,n))=0$ for all natural numbers $m,n$.



Let $F: mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 1$ if $m=n$ and $(m,n)mapsto 0$ if $mnot = n$.



Visually for these examples, we can think of



0,0,0, ...



0,0,0, ...



...



...



0,0,0, ...



...



and



1,0,0, ...



0,1,0, ...



0,0,1, ...



...



...



0,0,0, ...



...



where the locations represent elements in $mathbb{N}times mathbb{N}$, i.e.



(1,1), (1,2), (1,3), ...



(2,1), (2,2), (2,3), ...



...



...



($m$,1), ($m$,2), ... , ($m$,$n-1$), ($m$, $n$), ...



...



and the values that appear in those locations represent the element of ${0,1}$ which $F$ maps its location to.






share|cite|improve this answer









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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

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    2












    $begingroup$

    A function $F:mathbb{N}times mathbb{N} to {0,1}$ (“from $mathbb{N}times mathbb{N}$ to ${0,1}$”) assigns to each element of the set $mathbb{N}times mathbb{N}$, ordered pairs of integers, an element in the set {0,1}, either 0 or 1. So if $m$ and $n$ are natural numbers, $F((m,n))=0$ or $F((m,n))=1$.



    Some examples: let $F:mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 0$. On the left of “$to$”, we have the domain of our function. On the right, we have our codomain. On the left of “$mapsto$”, we have some arbitrary element in the domain. On the right, we have the explicit element in the codomain that $(m,n)$ will map to. So, in particular, $F((m,n))=0$ for all natural numbers $m,n$.



    Let $F: mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 1$ if $m=n$ and $(m,n)mapsto 0$ if $mnot = n$.



    Visually for these examples, we can think of



    0,0,0, ...



    0,0,0, ...



    ...



    ...



    0,0,0, ...



    ...



    and



    1,0,0, ...



    0,1,0, ...



    0,0,1, ...



    ...



    ...



    0,0,0, ...



    ...



    where the locations represent elements in $mathbb{N}times mathbb{N}$, i.e.



    (1,1), (1,2), (1,3), ...



    (2,1), (2,2), (2,3), ...



    ...



    ...



    ($m$,1), ($m$,2), ... , ($m$,$n-1$), ($m$, $n$), ...



    ...



    and the values that appear in those locations represent the element of ${0,1}$ which $F$ maps its location to.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A function $F:mathbb{N}times mathbb{N} to {0,1}$ (“from $mathbb{N}times mathbb{N}$ to ${0,1}$”) assigns to each element of the set $mathbb{N}times mathbb{N}$, ordered pairs of integers, an element in the set {0,1}, either 0 or 1. So if $m$ and $n$ are natural numbers, $F((m,n))=0$ or $F((m,n))=1$.



      Some examples: let $F:mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 0$. On the left of “$to$”, we have the domain of our function. On the right, we have our codomain. On the left of “$mapsto$”, we have some arbitrary element in the domain. On the right, we have the explicit element in the codomain that $(m,n)$ will map to. So, in particular, $F((m,n))=0$ for all natural numbers $m,n$.



      Let $F: mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 1$ if $m=n$ and $(m,n)mapsto 0$ if $mnot = n$.



      Visually for these examples, we can think of



      0,0,0, ...



      0,0,0, ...



      ...



      ...



      0,0,0, ...



      ...



      and



      1,0,0, ...



      0,1,0, ...



      0,0,1, ...



      ...



      ...



      0,0,0, ...



      ...



      where the locations represent elements in $mathbb{N}times mathbb{N}$, i.e.



      (1,1), (1,2), (1,3), ...



      (2,1), (2,2), (2,3), ...



      ...



      ...



      ($m$,1), ($m$,2), ... , ($m$,$n-1$), ($m$, $n$), ...



      ...



      and the values that appear in those locations represent the element of ${0,1}$ which $F$ maps its location to.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        A function $F:mathbb{N}times mathbb{N} to {0,1}$ (“from $mathbb{N}times mathbb{N}$ to ${0,1}$”) assigns to each element of the set $mathbb{N}times mathbb{N}$, ordered pairs of integers, an element in the set {0,1}, either 0 or 1. So if $m$ and $n$ are natural numbers, $F((m,n))=0$ or $F((m,n))=1$.



        Some examples: let $F:mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 0$. On the left of “$to$”, we have the domain of our function. On the right, we have our codomain. On the left of “$mapsto$”, we have some arbitrary element in the domain. On the right, we have the explicit element in the codomain that $(m,n)$ will map to. So, in particular, $F((m,n))=0$ for all natural numbers $m,n$.



        Let $F: mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 1$ if $m=n$ and $(m,n)mapsto 0$ if $mnot = n$.



        Visually for these examples, we can think of



        0,0,0, ...



        0,0,0, ...



        ...



        ...



        0,0,0, ...



        ...



        and



        1,0,0, ...



        0,1,0, ...



        0,0,1, ...



        ...



        ...



        0,0,0, ...



        ...



        where the locations represent elements in $mathbb{N}times mathbb{N}$, i.e.



        (1,1), (1,2), (1,3), ...



        (2,1), (2,2), (2,3), ...



        ...



        ...



        ($m$,1), ($m$,2), ... , ($m$,$n-1$), ($m$, $n$), ...



        ...



        and the values that appear in those locations represent the element of ${0,1}$ which $F$ maps its location to.






        share|cite|improve this answer









        $endgroup$



        A function $F:mathbb{N}times mathbb{N} to {0,1}$ (“from $mathbb{N}times mathbb{N}$ to ${0,1}$”) assigns to each element of the set $mathbb{N}times mathbb{N}$, ordered pairs of integers, an element in the set {0,1}, either 0 or 1. So if $m$ and $n$ are natural numbers, $F((m,n))=0$ or $F((m,n))=1$.



        Some examples: let $F:mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 0$. On the left of “$to$”, we have the domain of our function. On the right, we have our codomain. On the left of “$mapsto$”, we have some arbitrary element in the domain. On the right, we have the explicit element in the codomain that $(m,n)$ will map to. So, in particular, $F((m,n))=0$ for all natural numbers $m,n$.



        Let $F: mathbb{N}times mathbb{N} to {0,1}$ be given by $(m,n)mapsto 1$ if $m=n$ and $(m,n)mapsto 0$ if $mnot = n$.



        Visually for these examples, we can think of



        0,0,0, ...



        0,0,0, ...



        ...



        ...



        0,0,0, ...



        ...



        and



        1,0,0, ...



        0,1,0, ...



        0,0,1, ...



        ...



        ...



        0,0,0, ...



        ...



        where the locations represent elements in $mathbb{N}times mathbb{N}$, i.e.



        (1,1), (1,2), (1,3), ...



        (2,1), (2,2), (2,3), ...



        ...



        ...



        ($m$,1), ($m$,2), ... , ($m$,$n-1$), ($m$, $n$), ...



        ...



        and the values that appear in those locations represent the element of ${0,1}$ which $F$ maps its location to.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 4:58









        John BJohn B

        1766




        1766






























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