Well ordering and maximal chains in power set
$begingroup$
Let $M$ be a set and "$le$" a well-ordering of $M$. For $x in M$ define:
$$ M_{le x} := { y in M vert y le x } $$
The map
$$ f : M to mathcal{P}(M) , x mapsto M_{le x}$$
is injective and has the following property:
$$ x le y Leftrightarrow f(x) subseteq f(y)$$
Furthermore the set ${ emptyset } cup f(M) cup {M}$ is a maximal chain in $mathcal{P}(M)$ (ordered by "$subseteq$").
My question is: Does every maximal chain in $mathcal{P}(M)$ derive this way? Is there a bijection between the well-orderings of $M$ and the maximal chains in its power set? I was not yet able to proof that.
I would be super glad if this was even provable without the axiom of choice.
This would give a short proof of the well-ordering theorem by Hausdorffs maximal principle. Also, if $(M,le)$ is a partial ordered set and $K$ is a maximal chain in $mathcal{P}(M)$, I think that $f^{-1}(({emptyset} cup f(M) cup {M} ) cap K)$ (where $f$ is now defined via the partial ordering "$le$") is a maximal chain in $M$. So this would give a short proof of the maximal principle via well-ordering theorem. What do you think? I hope there is no obvious mistake...
Edit: Well, there was an obvious mistake and as you pointed out, maximal chains in $mathcal{P}(M)$ really dont need to relate to well-orderings of $M$. My original goal was to characterize the statement "$M$ is well-ordable" to a statement like "the maximal principle applies to $M$", whatever this should mean. I know the proof about maximal elements/chains in the set of partial well-orderings (well-orderings on subsets of $M$) being equivalent to well-orderings of $M$ but this doesnt seem useful to construct a maximal chain in $M$ if $M$ is partial ordered.
order-theory well-orders
asked Dec 22 '18 at 2:13
LucinaLucina
655
$endgroup$
|
show 2 more comments
$begingroup$
Let $M$ be a set and "$le$" a well-ordering of $M$. For $x in M$ define:
$$ M_{le x} := { y in M vert y le x } $$
The map
$$ f : M to mathcal{P}(M) , x mapsto M_{le x}$$
is injective and has the following property:
$$ x le y Leftrightarrow f(x) subseteq f(y)$$
Furthermore the set ${ emptyset } cup f(M) cup {M}$ is a maximal chain in $mathcal{P}(M)$ (ordered by "$subseteq$").
My question is: Does every maximal chain in $mathcal{P}(M)$ derive this way? Is there a bijection between the well-orderings of $M$ and the maximal chains in its power set? I was not yet able to proof that.
I would be super glad if this was even provable without the axiom of choice.
This would give a short proof of the well-ordering theorem by Hausdorffs maximal principle. Also, if $(M,le)$ is a partial ordered set and $K$ is a maximal chain in $mathcal{P}(M)$, I think that $f^{-1}(({emptyset} cup f(M) cup {M} ) cap K)$ (where $f$ is now defined via the partial ordering "$le$") is a maximal chain in $M$. So this would give a short proof of the maximal principle via well-ordering theorem. What do you think? I hope there is no obvious mistake...
Edit: Well, there was an obvious mistake and as you pointed out, maximal chains in $mathcal{P}(M)$ really dont need to relate to well-orderings of $M$. My original goal was to characterize the statement "$M$ is well-ordable" to a statement like "the maximal principle applies to $M$", whatever this should mean. I know the proof about maximal elements/chains in the set of partial well-orderings (well-orderings on subsets of $M$) being equivalent to well-orderings of $M$ but this doesnt seem useful to construct a maximal chain in $M$ if $M$ is partial ordered.
order-theory well-orders
asked Dec 22 '18 at 2:13
LucinaLucina
655
$endgroup$
$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
1
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14
|
show 2 more comments
$begingroup$
Let $M$ be a set and "$le$" a well-ordering of $M$. For $x in M$ define:
$$ M_{le x} := { y in M vert y le x } $$
The map
$$ f : M to mathcal{P}(M) , x mapsto M_{le x}$$
is injective and has the following property:
$$ x le y Leftrightarrow f(x) subseteq f(y)$$
Furthermore the set ${ emptyset } cup f(M) cup {M}$ is a maximal chain in $mathcal{P}(M)$ (ordered by "$subseteq$").
My question is: Does every maximal chain in $mathcal{P}(M)$ derive this way? Is there a bijection between the well-orderings of $M$ and the maximal chains in its power set? I was not yet able to proof that.
I would be super glad if this was even provable without the axiom of choice.
This would give a short proof of the well-ordering theorem by Hausdorffs maximal principle. Also, if $(M,le)$ is a partial ordered set and $K$ is a maximal chain in $mathcal{P}(M)$, I think that $f^{-1}(({emptyset} cup f(M) cup {M} ) cap K)$ (where $f$ is now defined via the partial ordering "$le$") is a maximal chain in $M$. So this would give a short proof of the maximal principle via well-ordering theorem. What do you think? I hope there is no obvious mistake...
Edit: Well, there was an obvious mistake and as you pointed out, maximal chains in $mathcal{P}(M)$ really dont need to relate to well-orderings of $M$. My original goal was to characterize the statement "$M$ is well-ordable" to a statement like "the maximal principle applies to $M$", whatever this should mean. I know the proof about maximal elements/chains in the set of partial well-orderings (well-orderings on subsets of $M$) being equivalent to well-orderings of $M$ but this doesnt seem useful to construct a maximal chain in $M$ if $M$ is partial ordered.
order-theory well-orders
asked Dec 22 '18 at 2:13
LucinaLucina
655
$endgroup$
Let $M$ be a set and "$le$" a well-ordering of $M$. For $x in M$ define:
$$ M_{le x} := { y in M vert y le x } $$
The map
$$ f : M to mathcal{P}(M) , x mapsto M_{le x}$$
is injective and has the following property:
$$ x le y Leftrightarrow f(x) subseteq f(y)$$
Furthermore the set ${ emptyset } cup f(M) cup {M}$ is a maximal chain in $mathcal{P}(M)$ (ordered by "$subseteq$").
My question is: Does every maximal chain in $mathcal{P}(M)$ derive this way? Is there a bijection between the well-orderings of $M$ and the maximal chains in its power set? I was not yet able to proof that.
I would be super glad if this was even provable without the axiom of choice.
This would give a short proof of the well-ordering theorem by Hausdorffs maximal principle. Also, if $(M,le)$ is a partial ordered set and $K$ is a maximal chain in $mathcal{P}(M)$, I think that $f^{-1}(({emptyset} cup f(M) cup {M} ) cap K)$ (where $f$ is now defined via the partial ordering "$le$") is a maximal chain in $M$. So this would give a short proof of the maximal principle via well-ordering theorem. What do you think? I hope there is no obvious mistake...
Edit: Well, there was an obvious mistake and as you pointed out, maximal chains in $mathcal{P}(M)$ really dont need to relate to well-orderings of $M$. My original goal was to characterize the statement "$M$ is well-ordable" to a statement like "the maximal principle applies to $M$", whatever this should mean. I know the proof about maximal elements/chains in the set of partial well-orderings (well-orderings on subsets of $M$) being equivalent to well-orderings of $M$ but this doesnt seem useful to construct a maximal chain in $M$ if $M$ is partial ordered.
order-theory well-orders
order-theory well-orders
asked Dec 22 '18 at 2:13
LucinaLucina
655
asked Dec 22 '18 at 2:13
LucinaLucina
655
edited Dec 23 '18 at 8:17
Lucina
asked Dec 22 '18 at 2:13
LucinaLucina
655
asked Dec 22 '18 at 2:13
LucinaLucina
655
asked Dec 22 '18 at 2:13
LucinaLucina
655
655
$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
1
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14
|
show 2 more comments
$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
1
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14
$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
1
1
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
No. Indeed, let $Asubsetmathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
Your construction works for any total order on $M,$ not just well-orderings.
And given a maximal chain in $P(M),$ you get a total order on $M.$ These are in 1-1 correspondence.
Start of Proof: Let $C$ be a maximal chain in $P(M).$ Then $emptyset, Min C,$ because the chain wouldn't be maximal if they weren't.
If $x,yin M$ we say $xleq_C y$ if and only if $forall Sin C$, if $yin Simplies xin S.$
To prove:
(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $xleq y$ and $yleq x,$ you'd have $x=y.$ This is due to the maximality of $C.$
(2) This is a total order.
The ultimate reason for this is that any maximal chain in $P(M)$ is closed under arbitrary unions and intersections. So the union $U_x$ of the elements of $C$ which do not contain $x$ and the intersection $V_x$ of all elements of $C$ which do contain $x$ are both in $C$, and there are no elements of $C$ strictly between $U_x$ and $V_x.$
For (1), this means that if $xleq y$ and $yleq x$ then $yin U_x$ and $ynotin V_x.$ But this means that $U_xcup{y}$ is strictly between $U_x$ and $V_x.$
For (2), this means that for $x.yin M,$ either $U_xsubseteq U_y$ or $U_ysubseteq U_x$, since $U_x,U_yin C$ and $C$ is a chain. But $U_xsubseteq U_y$ means $xleq y$ so either $xleq_C y$ or $yleq_C x.$
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
No. Indeed, let $Asubsetmathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
No. Indeed, let $Asubsetmathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
$endgroup$
add a comment |
$begingroup$
No. Indeed, let $Asubsetmathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
$endgroup$
No. Indeed, let $Asubsetmathcal{P}(M)$ be any chain that is not well-ordered (such a chain exists as long as $M$ is infinite; for instance you can just take an infinite descending sequence where you remove one element at a time). By Zorn's lemma, $A$ is contained in some maximal chain $B$ of $mathcal{P}(M)$, which is not well-ordered since it contains $A$. Since all the chains you describe are well-ordered, this $B$ cannot be of that form.
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
answered Dec 22 '18 at 2:21
Eric WofseyEric Wofsey
191k14216349
191k14216349
add a comment |
add a comment |
$begingroup$
Your construction works for any total order on $M,$ not just well-orderings.
And given a maximal chain in $P(M),$ you get a total order on $M.$ These are in 1-1 correspondence.
Start of Proof: Let $C$ be a maximal chain in $P(M).$ Then $emptyset, Min C,$ because the chain wouldn't be maximal if they weren't.
If $x,yin M$ we say $xleq_C y$ if and only if $forall Sin C$, if $yin Simplies xin S.$
To prove:
(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $xleq y$ and $yleq x,$ you'd have $x=y.$ This is due to the maximality of $C.$
(2) This is a total order.
The ultimate reason for this is that any maximal chain in $P(M)$ is closed under arbitrary unions and intersections. So the union $U_x$ of the elements of $C$ which do not contain $x$ and the intersection $V_x$ of all elements of $C$ which do contain $x$ are both in $C$, and there are no elements of $C$ strictly between $U_x$ and $V_x.$
For (1), this means that if $xleq y$ and $yleq x$ then $yin U_x$ and $ynotin V_x.$ But this means that $U_xcup{y}$ is strictly between $U_x$ and $V_x.$
For (2), this means that for $x.yin M,$ either $U_xsubseteq U_y$ or $U_ysubseteq U_x$, since $U_x,U_yin C$ and $C$ is a chain. But $U_xsubseteq U_y$ means $xleq y$ so either $xleq_C y$ or $yleq_C x.$
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Your construction works for any total order on $M,$ not just well-orderings.
And given a maximal chain in $P(M),$ you get a total order on $M.$ These are in 1-1 correspondence.
Start of Proof: Let $C$ be a maximal chain in $P(M).$ Then $emptyset, Min C,$ because the chain wouldn't be maximal if they weren't.
If $x,yin M$ we say $xleq_C y$ if and only if $forall Sin C$, if $yin Simplies xin S.$
To prove:
(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $xleq y$ and $yleq x,$ you'd have $x=y.$ This is due to the maximality of $C.$
(2) This is a total order.
The ultimate reason for this is that any maximal chain in $P(M)$ is closed under arbitrary unions and intersections. So the union $U_x$ of the elements of $C$ which do not contain $x$ and the intersection $V_x$ of all elements of $C$ which do contain $x$ are both in $C$, and there are no elements of $C$ strictly between $U_x$ and $V_x.$
For (1), this means that if $xleq y$ and $yleq x$ then $yin U_x$ and $ynotin V_x.$ But this means that $U_xcup{y}$ is strictly between $U_x$ and $V_x.$
For (2), this means that for $x.yin M,$ either $U_xsubseteq U_y$ or $U_ysubseteq U_x$, since $U_x,U_yin C$ and $C$ is a chain. But $U_xsubseteq U_y$ means $xleq y$ so either $xleq_C y$ or $yleq_C x.$
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
add a comment |
$begingroup$
Your construction works for any total order on $M,$ not just well-orderings.
And given a maximal chain in $P(M),$ you get a total order on $M.$ These are in 1-1 correspondence.
Start of Proof: Let $C$ be a maximal chain in $P(M).$ Then $emptyset, Min C,$ because the chain wouldn't be maximal if they weren't.
If $x,yin M$ we say $xleq_C y$ if and only if $forall Sin C$, if $yin Simplies xin S.$
To prove:
(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $xleq y$ and $yleq x,$ you'd have $x=y.$ This is due to the maximality of $C.$
(2) This is a total order.
The ultimate reason for this is that any maximal chain in $P(M)$ is closed under arbitrary unions and intersections. So the union $U_x$ of the elements of $C$ which do not contain $x$ and the intersection $V_x$ of all elements of $C$ which do contain $x$ are both in $C$, and there are no elements of $C$ strictly between $U_x$ and $V_x.$
For (1), this means that if $xleq y$ and $yleq x$ then $yin U_x$ and $ynotin V_x.$ But this means that $U_xcup{y}$ is strictly between $U_x$ and $V_x.$
For (2), this means that for $x.yin M,$ either $U_xsubseteq U_y$ or $U_ysubseteq U_x$, since $U_x,U_yin C$ and $C$ is a chain. But $U_xsubseteq U_y$ means $xleq y$ so either $xleq_C y$ or $yleq_C x.$
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
$endgroup$
Your construction works for any total order on $M,$ not just well-orderings.
And given a maximal chain in $P(M),$ you get a total order on $M.$ These are in 1-1 correspondence.
Start of Proof: Let $C$ be a maximal chain in $P(M).$ Then $emptyset, Min C,$ because the chain wouldn't be maximal if they weren't.
If $x,yin M$ we say $xleq_C y$ if and only if $forall Sin C$, if $yin Simplies xin S.$
To prove:
(1) This is a partial order. Reflexivity and transitivity are easy. The only hard part is proving if $xleq y$ and $yleq x,$ you'd have $x=y.$ This is due to the maximality of $C.$
(2) This is a total order.
The ultimate reason for this is that any maximal chain in $P(M)$ is closed under arbitrary unions and intersections. So the union $U_x$ of the elements of $C$ which do not contain $x$ and the intersection $V_x$ of all elements of $C$ which do contain $x$ are both in $C$, and there are no elements of $C$ strictly between $U_x$ and $V_x.$
For (1), this means that if $xleq y$ and $yleq x$ then $yin U_x$ and $ynotin V_x.$ But this means that $U_xcup{y}$ is strictly between $U_x$ and $V_x.$
For (2), this means that for $x.yin M,$ either $U_xsubseteq U_y$ or $U_ysubseteq U_x$, since $U_x,U_yin C$ and $C$ is a chain. But $U_xsubseteq U_y$ means $xleq y$ so either $xleq_C y$ or $yleq_C x.$
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
edited Dec 22 '18 at 19:13
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
answered Dec 22 '18 at 3:36
Thomas AndrewsThomas Andrews
130k12147298
130k12147298
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$begingroup$
As for your claim at the end, why would $f^{-1}(f(M)cap K)$ be maximal? You can find counterexamples to that even when $M$ has just two elements (try it!).
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:26
$begingroup$
Sorry, I wanted to write "$f^{-1}(({ emptyset} cup f(M) cup {M}) cap K)$". It will be corrected in the post.
$endgroup$
– Lucina
Dec 22 '18 at 2:33
$begingroup$
If $M$ is a countably infinite set, then the chains of the sort you describe are countable chains, but there are also uncountable chains in $mathcal P(M)$. For instance, the set of lower Dedekind cuts is an uncountable chain in the power set of the rational numbers.
$endgroup$
– bof
Dec 22 '18 at 2:41
$begingroup$
@Lucina: That doesn't help. The point is that maximality of $K$ in no way guarantees that $f(M)cap K$ is large, since most elements of $f(M)$ could be incomparable to elements of $K$. If you don't see this, I recommend actually thinking through what could happen if $M$ has two elements.
$endgroup$
– Eric Wofsey
Dec 22 '18 at 2:45
1
$begingroup$
Are you possibly confusing "well-ordering" with "total ordering?" Because your construction can be done with any total order on $M,$ and any maximal chain on $P(M)$ is a total order on $M.$
$endgroup$
– Thomas Andrews
Dec 22 '18 at 3:14