If $X$ is a set of ordinals then so is $bigcup X$Ordinals with the same cardinalityRelation between two distinct class of ordinals$bigcup X$ = $sup(X)$ for a set $X$ of ordinalsElementary Set Theory: Ordinals, Well Orderings and Isomorphismsconfusion about the definition of cardinalityIf $X$ is well-ordered then $Xcup left infty right$ is well-ordered.If $x$ is a set of ordinals, then $bigcup x=alpha$. Why is $xsubseteq alpha+1$ rather than $alpha$Class of Ordinals $On$ not a SetSuppose that $X$ is a set of ordinals. Then there exists an ordinal $betanotin X$Fundemental Properties of Ordinals
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If $X$ is a set of ordinals then so is $bigcup X$
Ordinals with the same cardinalityRelation between two distinct class of ordinals$bigcup X$ = $sup(X)$ for a set $X$ of ordinalsElementary Set Theory: Ordinals, Well Orderings and Isomorphismsconfusion about the definition of cardinalityIf $X$ is well-ordered then $Xcup left infty right$ is well-ordered.If $x$ is a set of ordinals, then $bigcup x=alpha$. Why is $xsubseteq alpha+1$ rather than $alpha$Class of Ordinals $On$ not a SetSuppose that $X$ is a set of ordinals. Then there exists an ordinal $betanotin X$Fundemental Properties of Ordinals
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Theorem. If $X$ is a set of ordinals then so is $bigcup X.$
Claim 1. $bigcup X$ is well-ordered.
Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.
Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$
Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$
Proof of theorem. By the Claim 1 and the Claim 2.
May you check my proof? Thanks...
elementary-set-theory proof-writing
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
$endgroup$
add a comment |
$begingroup$
Theorem. If $X$ is a set of ordinals then so is $bigcup X.$
Claim 1. $bigcup X$ is well-ordered.
Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.
Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$
Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$
Proof of theorem. By the Claim 1 and the Claim 2.
May you check my proof? Thanks...
elementary-set-theory proof-writing
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
$endgroup$
$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
1
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02
add a comment |
$begingroup$
Theorem. If $X$ is a set of ordinals then so is $bigcup X.$
Claim 1. $bigcup X$ is well-ordered.
Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.
Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$
Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$
Proof of theorem. By the Claim 1 and the Claim 2.
May you check my proof? Thanks...
elementary-set-theory proof-writing
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
$endgroup$
Theorem. If $X$ is a set of ordinals then so is $bigcup X.$
Claim 1. $bigcup X$ is well-ordered.
Proof of Claim 1. Let $Ysubseteqbigcup_Iin XI.$ There is a $Iin X$ such that $Ysubseteq X.$ Since $X$ is well-ordered, $Y$ does have a least element, so $bigcup_Iin XX$ is well-ordered.
Claim 2. If $xinbigcup X$, then $xsubseteqbigcup X.$
Proof of Claim 2. Let $xinbigcup X.$ So there is a $Iin X$ such that $xin X.$ Since $X$ is an ordinal, $xsubseteq Xsubseteq bigcup X$, so $xsubseteqbigcup X.$
Proof of theorem. By the Claim 1 and the Claim 2.
May you check my proof? Thanks...
elementary-set-theory proof-writing
elementary-set-theory proof-writing
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
asked Mar 17 at 21:28
PozcuKushimotoStreetPozcuKushimotoStreet
1,408923
1,408923
$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
1
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02
add a comment |
$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
1
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02
$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
1
1
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02
add a comment |
1 Answer
1
active
oldest
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$begingroup$
So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.
However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.
Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.
The proof of claim 2 is correct.
answered Mar 18 at 14:56
BerciBerci
61.8k23674
$endgroup$
add a comment |
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$begingroup$
So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.
However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.
Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.
The proof of claim 2 is correct.
answered Mar 18 at 14:56
BerciBerci
61.8k23674
$endgroup$
add a comment |
$begingroup$
So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.
However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.
Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.
The proof of claim 2 is correct.
answered Mar 18 at 14:56
BerciBerci
61.8k23674
$endgroup$
add a comment |
$begingroup$
So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.
However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.
Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.
The proof of claim 2 is correct.
answered Mar 18 at 14:56
BerciBerci
61.8k23674
$endgroup$
So, for claim 1, let $Ysubseteq bigcup X$, then it is not true in general that $Ysubseteq I$ for some $Iin X$.
However, it's true for any element $y$ of $Y$: $yin I$ with some $Iin X$.
Now we can take the smallest element $y_0$ in $Ycap I, subseteq I$, and prove it's the smallest element of $Y$.
The proof of claim 2 is correct.
answered Mar 18 at 14:56
BerciBerci
61.8k23674
answered Mar 18 at 14:56
BerciBerci
61.8k23674
answered Mar 18 at 14:56
BerciBerci
61.8k23674
answered Mar 18 at 14:56
BerciBerci
61.8k23674
61.8k23674
add a comment |
add a comment |
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$begingroup$
You got confused by the notation. You might use like $bigcup mathcal X=bigcupXinmathcal XX$. Other than that, proof of claim 2 is correct.
$endgroup$
– Berci
Mar 17 at 21:38
$begingroup$
@Berci I couldn't understand your notation. Can you explain in detail?
$endgroup$
– PozcuKushimotoStreet
Mar 17 at 21:46
1
$begingroup$
Only that you wrote $bigcup_Iin XX$ instead of $bigcup_Iin XI$.
$endgroup$
– Berci
Mar 18 at 12:02