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:-
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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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:-
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
add a comment |
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:-
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
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:-
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
general-topology ordinals
edited Sep 11 at 2:24
Henning Makholm
234k16299532
234k16299532
asked Sep 11 at 1:38
Math geek
35118
35118
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1 Answer
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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.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
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.
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.
answered Sep 11 at 2:21
Henning Makholm
234k16299532
234k16299532
add a comment |
add a comment |
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