Proving Hausdorff maximality principle without Choice












1














I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for
every partial order (P, ≤) ∈ V with the property that there exists a well-
order ≺ on the underlying set P, there is an inclusion-maximal chain X
in (P, ≤). How could I prove it using only ZF and not Choice?






set-theory axiom-of-choice







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  • 1




    Do you happen to have heard that in problem set 5?
    – Alessandro Codenotti
    Nov 14 '18 at 9:18
















1














I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for
every partial order (P, ≤) ∈ V with the property that there exists a well-
order ≺ on the underlying set P, there is an inclusion-maximal chain X
in (P, ≤). How could I prove it using only ZF and not Choice?






set-theory axiom-of-choice







share|cite|improve this question




















  • 1




    Do you happen to have heard that in problem set 5?
    – Alessandro Codenotti
    Nov 14 '18 at 9:18














1












1








1







I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for
every partial order (P, ≤) ∈ V with the property that there exists a well-
order ≺ on the underlying set P, there is an inclusion-maximal chain X
in (P, ≤). How could I prove it using only ZF and not Choice?






set-theory axiom-of-choice







share|cite|improve this question















I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for
every partial order (P, ≤) ∈ V with the property that there exists a well-
order ≺ on the underlying set P, there is an inclusion-maximal chain X
in (P, ≤). How could I prove it using only ZF and not Choice?






set-theory axiom-of-choice




set-theory axiom-of-choice






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 4 at 17:55









Andrés E. Caicedo

64.9k8158246




64.9k8158246










asked Nov 13 '18 at 18:13









P. GreweP. Grewe

61




61








  • 1




    Do you happen to have heard that in problem set 5?
    – Alessandro Codenotti
    Nov 14 '18 at 9:18














  • 1




    Do you happen to have heard that in problem set 5?
    – Alessandro Codenotti
    Nov 14 '18 at 9:18








1




1




Do you happen to have heard that in problem set 5?
– Alessandro Codenotti
Nov 14 '18 at 9:18




Do you happen to have heard that in problem set 5?
– Alessandro Codenotti
Nov 14 '18 at 9:18










1 Answer
1






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Recursively construct $(C_xi mid xi in mathrm{Ord})$ as follows:



Let $C_0 := emptyset$ and given $C_xi$ either $C_xi$ is a $subseteq$-maximal chain in $(P; le)$ in which case we stop the construction or otherwise
$$
C_{xi +1} := C_xi cup min_{prec} { p in P setminus C_xi mid forall c in C_xi colon p le c vee c le p }.
$$
For limit $lambda in mathrm{Ord}$, we let $C_{lambda} := bigcup_{xi < lambda} C_{xi}$.



It is easy to verify that each $C_xi$, if defined, is a chain through $(P; le)$ and that there is some $xi < H(P)$ such that $C_{xi}$ is a maximal chain.



(Here $H(P)$ is the least ordinal $alpha$ such that there is no injection $i colon alpha to P$. Since $P$ has a well-ordered, we have that $H(P) = mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)






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  • What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
    – Asaf Karagila♦
    Nov 13 '18 at 20:24






  • 1




    @AsafKaragila As you can see, I've mentioned that.
    – Stefan Mesken
    Nov 13 '18 at 20:25






  • 1




    As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
    – Asaf Karagila♦
    Nov 13 '18 at 20:26











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Recursively construct $(C_xi mid xi in mathrm{Ord})$ as follows:



Let $C_0 := emptyset$ and given $C_xi$ either $C_xi$ is a $subseteq$-maximal chain in $(P; le)$ in which case we stop the construction or otherwise
$$
C_{xi +1} := C_xi cup min_{prec} { p in P setminus C_xi mid forall c in C_xi colon p le c vee c le p }.
$$
For limit $lambda in mathrm{Ord}$, we let $C_{lambda} := bigcup_{xi < lambda} C_{xi}$.



It is easy to verify that each $C_xi$, if defined, is a chain through $(P; le)$ and that there is some $xi < H(P)$ such that $C_{xi}$ is a maximal chain.



(Here $H(P)$ is the least ordinal $alpha$ such that there is no injection $i colon alpha to P$. Since $P$ has a well-ordered, we have that $H(P) = mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)






share|cite|improve this answer























  • What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
    – Asaf Karagila♦
    Nov 13 '18 at 20:24






  • 1




    @AsafKaragila As you can see, I've mentioned that.
    – Stefan Mesken
    Nov 13 '18 at 20:25






  • 1




    As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
    – Asaf Karagila♦
    Nov 13 '18 at 20:26
















2














Recursively construct $(C_xi mid xi in mathrm{Ord})$ as follows:



Let $C_0 := emptyset$ and given $C_xi$ either $C_xi$ is a $subseteq$-maximal chain in $(P; le)$ in which case we stop the construction or otherwise
$$
C_{xi +1} := C_xi cup min_{prec} { p in P setminus C_xi mid forall c in C_xi colon p le c vee c le p }.
$$
For limit $lambda in mathrm{Ord}$, we let $C_{lambda} := bigcup_{xi < lambda} C_{xi}$.



It is easy to verify that each $C_xi$, if defined, is a chain through $(P; le)$ and that there is some $xi < H(P)$ such that $C_{xi}$ is a maximal chain.



(Here $H(P)$ is the least ordinal $alpha$ such that there is no injection $i colon alpha to P$. Since $P$ has a well-ordered, we have that $H(P) = mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)






share|cite|improve this answer























  • What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
    – Asaf Karagila♦
    Nov 13 '18 at 20:24






  • 1




    @AsafKaragila As you can see, I've mentioned that.
    – Stefan Mesken
    Nov 13 '18 at 20:25






  • 1




    As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
    – Asaf Karagila♦
    Nov 13 '18 at 20:26














2












2








2






Recursively construct $(C_xi mid xi in mathrm{Ord})$ as follows:



Let $C_0 := emptyset$ and given $C_xi$ either $C_xi$ is a $subseteq$-maximal chain in $(P; le)$ in which case we stop the construction or otherwise
$$
C_{xi +1} := C_xi cup min_{prec} { p in P setminus C_xi mid forall c in C_xi colon p le c vee c le p }.
$$
For limit $lambda in mathrm{Ord}$, we let $C_{lambda} := bigcup_{xi < lambda} C_{xi}$.



It is easy to verify that each $C_xi$, if defined, is a chain through $(P; le)$ and that there is some $xi < H(P)$ such that $C_{xi}$ is a maximal chain.



(Here $H(P)$ is the least ordinal $alpha$ such that there is no injection $i colon alpha to P$. Since $P$ has a well-ordered, we have that $H(P) = mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)






share|cite|improve this answer














Recursively construct $(C_xi mid xi in mathrm{Ord})$ as follows:



Let $C_0 := emptyset$ and given $C_xi$ either $C_xi$ is a $subseteq$-maximal chain in $(P; le)$ in which case we stop the construction or otherwise
$$
C_{xi +1} := C_xi cup min_{prec} { p in P setminus C_xi mid forall c in C_xi colon p le c vee c le p }.
$$
For limit $lambda in mathrm{Ord}$, we let $C_{lambda} := bigcup_{xi < lambda} C_{xi}$.



It is easy to verify that each $C_xi$, if defined, is a chain through $(P; le)$ and that there is some $xi < H(P)$ such that $C_{xi}$ is a maximal chain.



(Here $H(P)$ is the least ordinal $alpha$ such that there is no injection $i colon alpha to P$. Since $P$ has a well-ordered, we have that $H(P) = mathrm{card}(P)^+$ but for the sake of this proof it's actually more natural to think about it as $H(P)$ -- the Hartogs number of $P$.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 13 '18 at 20:12

























answered Nov 13 '18 at 20:01









Stefan MeskenStefan Mesken

14.5k32046




14.5k32046












  • What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
    – Asaf Karagila♦
    Nov 13 '18 at 20:24






  • 1




    @AsafKaragila As you can see, I've mentioned that.
    – Stefan Mesken
    Nov 13 '18 at 20:25






  • 1




    As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
    – Asaf Karagila♦
    Nov 13 '18 at 20:26


















  • What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
    – Asaf Karagila♦
    Nov 13 '18 at 20:24






  • 1




    @AsafKaragila As you can see, I've mentioned that.
    – Stefan Mesken
    Nov 13 '18 at 20:25






  • 1




    As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
    – Asaf Karagila♦
    Nov 13 '18 at 20:26
















What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
– Asaf Karagila♦
Nov 13 '18 at 20:24




What? Why? There is a well-ordering. You don't need to appeal to the Hartogs number at all.
– Asaf Karagila♦
Nov 13 '18 at 20:24




1




1




@AsafKaragila As you can see, I've mentioned that.
– Stefan Mesken
Nov 13 '18 at 20:25




@AsafKaragila As you can see, I've mentioned that.
– Stefan Mesken
Nov 13 '18 at 20:25




1




1




As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
– Asaf Karagila♦
Nov 13 '18 at 20:26




As you can see, I didn't actually read your answer. Just skimmed it a little bit... :P
– Asaf Karagila♦
Nov 13 '18 at 20:26


















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