The Hartogs number of $A$ exists for any set $A$
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The Hartogs number of $A$ exists for any set $A$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $H={(W,prec) in mathcal{P}(A) times mathcal{P}(Atimes A) mid prec text{ is a well-ordering on } W}$. It is well-known that every $(W,prec)in H$ is isomorphic to a unique ordinal.
By Axiom of Replacement, $h(A)={alpha in text{Ord} mid (alpha,<) text{ is isomorphic to some } (W,prec) in H}$ is a set. It follows that $h(A)$ is a set of ordinals.
Assume $gammainalphain h(A)$. Then there is an isomorphism $f$ between $(alpha,<)$ and some $(W,prec) in H$. Then $f restriction gamma$ is an isomorphism between $(gamma,<)$ and $(f[alpha],prec) in H$. Thus $gamma in H$. Hence, $h(A)$ is transitive.
To sum up, $h(A)$ is a set of ordinals and transitive. Thus $h(A)$ is an ordinal. Moreover, $h(A)$ is not equipotent to any subset of $A$. If not, there is a bijecton $g$ from $h(A)$ to some $W in mathcal{P}(A)$. We define a well-ordering $prec$ on $W$ by $w_1 prec w_2 iff g^{-1}(w_1) < g^{-1}(w_2)$. As such, $(h(A),<)$ is isomorphic to $(W,prec)$. Then $h(A)in h(A)$, which is a contradiction. This completes the proof.
proof-verification elementary-set-theory ordinals
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
add a comment |
$begingroup$
The Hartogs number of $A$ exists for any set $A$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $H={(W,prec) in mathcal{P}(A) times mathcal{P}(Atimes A) mid prec text{ is a well-ordering on } W}$. It is well-known that every $(W,prec)in H$ is isomorphic to a unique ordinal.
By Axiom of Replacement, $h(A)={alpha in text{Ord} mid (alpha,<) text{ is isomorphic to some } (W,prec) in H}$ is a set. It follows that $h(A)$ is a set of ordinals.
Assume $gammainalphain h(A)$. Then there is an isomorphism $f$ between $(alpha,<)$ and some $(W,prec) in H$. Then $f restriction gamma$ is an isomorphism between $(gamma,<)$ and $(f[alpha],prec) in H$. Thus $gamma in H$. Hence, $h(A)$ is transitive.
To sum up, $h(A)$ is a set of ordinals and transitive. Thus $h(A)$ is an ordinal. Moreover, $h(A)$ is not equipotent to any subset of $A$. If not, there is a bijecton $g$ from $h(A)$ to some $W in mathcal{P}(A)$. We define a well-ordering $prec$ on $W$ by $w_1 prec w_2 iff g^{-1}(w_1) < g^{-1}(w_2)$. As such, $(h(A),<)$ is isomorphic to $(W,prec)$. Then $h(A)in h(A)$, which is a contradiction. This completes the proof.
proof-verification elementary-set-theory ordinals
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
1
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25
add a comment |
$begingroup$
The Hartogs number of $A$ exists for any set $A$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $H={(W,prec) in mathcal{P}(A) times mathcal{P}(Atimes A) mid prec text{ is a well-ordering on } W}$. It is well-known that every $(W,prec)in H$ is isomorphic to a unique ordinal.
By Axiom of Replacement, $h(A)={alpha in text{Ord} mid (alpha,<) text{ is isomorphic to some } (W,prec) in H}$ is a set. It follows that $h(A)$ is a set of ordinals.
Assume $gammainalphain h(A)$. Then there is an isomorphism $f$ between $(alpha,<)$ and some $(W,prec) in H$. Then $f restriction gamma$ is an isomorphism between $(gamma,<)$ and $(f[alpha],prec) in H$. Thus $gamma in H$. Hence, $h(A)$ is transitive.
To sum up, $h(A)$ is a set of ordinals and transitive. Thus $h(A)$ is an ordinal. Moreover, $h(A)$ is not equipotent to any subset of $A$. If not, there is a bijecton $g$ from $h(A)$ to some $W in mathcal{P}(A)$. We define a well-ordering $prec$ on $W$ by $w_1 prec w_2 iff g^{-1}(w_1) < g^{-1}(w_2)$. As such, $(h(A),<)$ is isomorphic to $(W,prec)$. Then $h(A)in h(A)$, which is a contradiction. This completes the proof.
proof-verification elementary-set-theory ordinals
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
$endgroup$
The Hartogs number of $A$ exists for any set $A$.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Let $H={(W,prec) in mathcal{P}(A) times mathcal{P}(Atimes A) mid prec text{ is a well-ordering on } W}$. It is well-known that every $(W,prec)in H$ is isomorphic to a unique ordinal.
By Axiom of Replacement, $h(A)={alpha in text{Ord} mid (alpha,<) text{ is isomorphic to some } (W,prec) in H}$ is a set. It follows that $h(A)$ is a set of ordinals.
Assume $gammainalphain h(A)$. Then there is an isomorphism $f$ between $(alpha,<)$ and some $(W,prec) in H$. Then $f restriction gamma$ is an isomorphism between $(gamma,<)$ and $(f[alpha],prec) in H$. Thus $gamma in H$. Hence, $h(A)$ is transitive.
To sum up, $h(A)$ is a set of ordinals and transitive. Thus $h(A)$ is an ordinal. Moreover, $h(A)$ is not equipotent to any subset of $A$. If not, there is a bijecton $g$ from $h(A)$ to some $W in mathcal{P}(A)$. We define a well-ordering $prec$ on $W$ by $w_1 prec w_2 iff g^{-1}(w_1) < g^{-1}(w_2)$. As such, $(h(A),<)$ is isomorphic to $(W,prec)$. Then $h(A)in h(A)$, which is a contradiction. This completes the proof.
proof-verification elementary-set-theory ordinals
proof-verification elementary-set-theory ordinals
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
asked Dec 5 '18 at 2:57
Le Anh DungLe Anh Dung
1,0291521
1,0291521
1
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25
add a comment |
1
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25
1
1
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25
add a comment |
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1
$begingroup$
I see no problems with it.
$endgroup$
– Henno Brandsma
Dec 5 '18 at 3:27
$begingroup$
Thank you so much for your verification @HennoBrandsma :)
$endgroup$
– Le Anh Dung
Dec 5 '18 at 3:28
$begingroup$
Where do you think that your proof might contain a mistake?
$endgroup$
– Asaf Karagila♦
Dec 5 '18 at 7:25