Does $x in mathcal X:x>a land x<b+1 $ give $(a,b+1)$ or $(a,b]$?

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Ordinal numbers:- An ordinal number is a set $alpha$ with the following properties:



(1) If $x,y in alpha$, then either $xin y, yin x, $ or $y=x$



(2)If $y in alpha$ and $xin y$, then $xin alpha$.



Theorem H.10 There is an uncountable well-ordered set $Omega$ of ordinal numbers with maximal element $omega_1$ having the property if
$xin Omega$ and $xneq omega_1$, then $yin Omega:yleq x$ is
countable.




Definition of interval in the book is given by:-



enter image description here



enter image description here



Doubt:-
By the definition of the interval $x in mathcal X:x>a cap x in mathcal X:x<b+1 =(a,b+1)$. Am I correct? If there is no element between $b$ and $b+1$. I can agree with the notation in the textbook. Please help me.










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    down vote

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    Ordinal numbers:- An ordinal number is a set $alpha$ with the following properties:



    (1) If $x,y in alpha$, then either $xin y, yin x, $ or $y=x$



    (2)If $y in alpha$ and $xin y$, then $xin alpha$.



    Theorem H.10 There is an uncountable well-ordered set $Omega$ of ordinal numbers with maximal element $omega_1$ having the property if
    $xin Omega$ and $xneq omega_1$, then $yin Omega:yleq x$ is
    countable.




    Definition of interval in the book is given by:-



    enter image description here



    enter image description here



    Doubt:-
    By the definition of the interval $x in mathcal X:x>a cap x in mathcal X:x<b+1 =(a,b+1)$. Am I correct? If there is no element between $b$ and $b+1$. I can agree with the notation in the textbook. Please help me.










    share|cite|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      Ordinal numbers:- An ordinal number is a set $alpha$ with the following properties:



      (1) If $x,y in alpha$, then either $xin y, yin x, $ or $y=x$



      (2)If $y in alpha$ and $xin y$, then $xin alpha$.



      Theorem H.10 There is an uncountable well-ordered set $Omega$ of ordinal numbers with maximal element $omega_1$ having the property if
      $xin Omega$ and $xneq omega_1$, then $yin Omega:yleq x$ is
      countable.




      Definition of interval in the book is given by:-



      enter image description here



      enter image description here



      Doubt:-
      By the definition of the interval $x in mathcal X:x>a cap x in mathcal X:x<b+1 =(a,b+1)$. Am I correct? If there is no element between $b$ and $b+1$. I can agree with the notation in the textbook. Please help me.










      share|cite|improve this question
















      Ordinal numbers:- An ordinal number is a set $alpha$ with the following properties:



      (1) If $x,y in alpha$, then either $xin y, yin x, $ or $y=x$



      (2)If $y in alpha$ and $xin y$, then $xin alpha$.



      Theorem H.10 There is an uncountable well-ordered set $Omega$ of ordinal numbers with maximal element $omega_1$ having the property if
      $xin Omega$ and $xneq omega_1$, then $yin Omega:yleq x$ is
      countable.




      Definition of interval in the book is given by:-



      enter image description here



      enter image description here



      Doubt:-
      By the definition of the interval $x in mathcal X:x>a cap x in mathcal X:x<b+1 =(a,b+1)$. Am I correct? If there is no element between $b$ and $b+1$. I can agree with the notation in the textbook. Please help me.







      general-topology ordinals






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      edited Sep 11 at 2:24









      Henning Makholm

      234k16299532




      234k16299532










      asked Sep 11 at 1:38









      Math geek

      35118




      35118




















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          It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.



          (Recall that $b+1$ is by definition the set $bcupb$, and the standard ordering of ordinals is the same as ordering by $subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).



          Therefore the condition $x<b+1$ is the same as $xle b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.






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            active

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            1






            active

            oldest

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            oldest

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            active

            oldest

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            up vote
            3
            down vote



            accepted










            It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.



            (Recall that $b+1$ is by definition the set $bcupb$, and the standard ordering of ordinals is the same as ordering by $subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).



            Therefore the condition $x<b+1$ is the same as $xle b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.






            share|cite|improve this answer
























              up vote
              3
              down vote



              accepted










              It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.



              (Recall that $b+1$ is by definition the set $bcupb$, and the standard ordering of ordinals is the same as ordering by $subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).



              Therefore the condition $x<b+1$ is the same as $xle b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.






              share|cite|improve this answer






















                up vote
                3
                down vote



                accepted







                up vote
                3
                down vote



                accepted






                It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.



                (Recall that $b+1$ is by definition the set $bcupb$, and the standard ordering of ordinals is the same as ordering by $subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).



                Therefore the condition $x<b+1$ is the same as $xle b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.






                share|cite|improve this answer












                It is important here that we're talking about sets of ordinals, so we know there is no possible element strictly between $b$ and $b+1$.



                (Recall that $b+1$ is by definition the set $bcupb$, and the standard ordering of ordinals is the same as ordering by $subseteq$, so there can't be any ordinal strictly between $b$ and $b+1$ because $b+1$ has only one additional element).



                Therefore the condition $x<b+1$ is the same as $xle b$, so $(a,b+1)$ and $(a,b]$ are two ways of writing the same set of ordinals.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 11 at 2:21









                Henning Makholm

                234k16299532




                234k16299532



























                     

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