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Prove that lower bounds of upper bounds of a subset is a closure operator on a lattice.
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Let $(L, le)$ be a poset regarded as a lattice (s.t. $forall a, b in L$ both $sup({a,b})$ and $inf({a,b})$ exist in L).
Let $A subseteq L$. We denote
$$U(A):= { bin L | forall ain A: a leq b} \
L(A):= { bin L | forall ain A: b leq a}$$
sets of all upper and lower bounds of $A$.
I need to prove that map $C(A):= L(U(A))$ defines a closure operator on $L$.
That is I need to prove 3 properties:
1) $Asubseteq C(A)$
2) $C(C(A)) = C(A)$
3) $Asubseteq B implies C(A) subseteq C(B)$
My try:
1) $bin A implies forall ain U(A): bleq a implies b in L(U(A)) implies A subseteq L(U(A)) = C(A)$
2) ?
3) We have property: $A subseteq B implies U(B) subseteq U(A)$ and $L(B) subseteq L(A)$.
Thus
$ Asubseteq B implies U(B) subseteq U(A) implies L(U(A)) subseteq L(U(B))$
I'm strugling to prove idempotency property. I guess to prove it I need to show that $$U(A) = U(L(U(A)))$$ but I don't know how to do it formally.
Question: are these proofs valid and how do I prove idempotency?
Edit:
2) $U(L(U(A))) subseteq U(A)$ from properties (1) and (3).
$$a in U(A) implies forall b in L(U(A)): b leq a implies a in U(L(U(A))) implies U(A) subseteq U(L(U(A)))$$
That is equality holds, thus $C$ is idempotent.
Thanks everyone!
order-theory lattice-orders
asked yesterday
Sergey Dylda
1176
add a comment |
up vote
0
down vote
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Let $(L, le)$ be a poset regarded as a lattice (s.t. $forall a, b in L$ both $sup({a,b})$ and $inf({a,b})$ exist in L).
Let $A subseteq L$. We denote
$$U(A):= { bin L | forall ain A: a leq b} \
L(A):= { bin L | forall ain A: b leq a}$$
sets of all upper and lower bounds of $A$.
I need to prove that map $C(A):= L(U(A))$ defines a closure operator on $L$.
That is I need to prove 3 properties:
1) $Asubseteq C(A)$
2) $C(C(A)) = C(A)$
3) $Asubseteq B implies C(A) subseteq C(B)$
My try:
1) $bin A implies forall ain U(A): bleq a implies b in L(U(A)) implies A subseteq L(U(A)) = C(A)$
2) ?
3) We have property: $A subseteq B implies U(B) subseteq U(A)$ and $L(B) subseteq L(A)$.
Thus
$ Asubseteq B implies U(B) subseteq U(A) implies L(U(A)) subseteq L(U(B))$
I'm strugling to prove idempotency property. I guess to prove it I need to show that $$U(A) = U(L(U(A)))$$ but I don't know how to do it formally.
Question: are these proofs valid and how do I prove idempotency?
Edit:
2) $U(L(U(A))) subseteq U(A)$ from properties (1) and (3).
$$a in U(A) implies forall b in L(U(A)): b leq a implies a in U(L(U(A))) implies U(A) subseteq U(L(U(A)))$$
That is equality holds, thus $C$ is idempotent.
Thanks everyone!
order-theory lattice-orders
asked yesterday
Sergey Dylda
1176
1
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(L, le)$ be a poset regarded as a lattice (s.t. $forall a, b in L$ both $sup({a,b})$ and $inf({a,b})$ exist in L).
Let $A subseteq L$. We denote
$$U(A):= { bin L | forall ain A: a leq b} \
L(A):= { bin L | forall ain A: b leq a}$$
sets of all upper and lower bounds of $A$.
I need to prove that map $C(A):= L(U(A))$ defines a closure operator on $L$.
That is I need to prove 3 properties:
1) $Asubseteq C(A)$
2) $C(C(A)) = C(A)$
3) $Asubseteq B implies C(A) subseteq C(B)$
My try:
1) $bin A implies forall ain U(A): bleq a implies b in L(U(A)) implies A subseteq L(U(A)) = C(A)$
2) ?
3) We have property: $A subseteq B implies U(B) subseteq U(A)$ and $L(B) subseteq L(A)$.
Thus
$ Asubseteq B implies U(B) subseteq U(A) implies L(U(A)) subseteq L(U(B))$
I'm strugling to prove idempotency property. I guess to prove it I need to show that $$U(A) = U(L(U(A)))$$ but I don't know how to do it formally.
Question: are these proofs valid and how do I prove idempotency?
Edit:
2) $U(L(U(A))) subseteq U(A)$ from properties (1) and (3).
$$a in U(A) implies forall b in L(U(A)): b leq a implies a in U(L(U(A))) implies U(A) subseteq U(L(U(A)))$$
That is equality holds, thus $C$ is idempotent.
Thanks everyone!
order-theory lattice-orders
asked yesterday
Sergey Dylda
1176
Let $(L, le)$ be a poset regarded as a lattice (s.t. $forall a, b in L$ both $sup({a,b})$ and $inf({a,b})$ exist in L).
Let $A subseteq L$. We denote
$$U(A):= { bin L | forall ain A: a leq b} \
L(A):= { bin L | forall ain A: b leq a}$$
sets of all upper and lower bounds of $A$.
I need to prove that map $C(A):= L(U(A))$ defines a closure operator on $L$.
That is I need to prove 3 properties:
1) $Asubseteq C(A)$
2) $C(C(A)) = C(A)$
3) $Asubseteq B implies C(A) subseteq C(B)$
My try:
1) $bin A implies forall ain U(A): bleq a implies b in L(U(A)) implies A subseteq L(U(A)) = C(A)$
2) ?
3) We have property: $A subseteq B implies U(B) subseteq U(A)$ and $L(B) subseteq L(A)$.
Thus
$ Asubseteq B implies U(B) subseteq U(A) implies L(U(A)) subseteq L(U(B))$
I'm strugling to prove idempotency property. I guess to prove it I need to show that $$U(A) = U(L(U(A)))$$ but I don't know how to do it formally.
Question: are these proofs valid and how do I prove idempotency?
Edit:
2) $U(L(U(A))) subseteq U(A)$ from properties (1) and (3).
$$a in U(A) implies forall b in L(U(A)): b leq a implies a in U(L(U(A))) implies U(A) subseteq U(L(U(A)))$$
That is equality holds, thus $C$ is idempotent.
Thanks everyone!
order-theory lattice-orders
order-theory lattice-orders
asked yesterday
Sergey Dylda
1176
asked yesterday
Sergey Dylda
1176
edited yesterday
asked yesterday
Sergey Dylda
1176
asked yesterday
Sergey Dylda
1176
asked yesterday
Sergey Dylda
1176
1176
1
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday
add a comment |
1
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday
1
1
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
As Kevin Long wrote in a comment, you already have that $U(A) subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A subseteq B$ implies $U(B) subseteq U(A)$,
$$A subseteq L(U(A)) Longrightarrow U(L(U(A))) subseteq U(A).$$
Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.
answered yesterday
amrsa
3,4052518
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As Kevin Long wrote in a comment, you already have that $U(A) subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A subseteq B$ implies $U(B) subseteq U(A)$,
$$A subseteq L(U(A)) Longrightarrow U(L(U(A))) subseteq U(A).$$
Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.
answered yesterday
amrsa
3,4052518
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
add a comment |
up vote
1
down vote
accepted
As Kevin Long wrote in a comment, you already have that $U(A) subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A subseteq B$ implies $U(B) subseteq U(A)$,
$$A subseteq L(U(A)) Longrightarrow U(L(U(A))) subseteq U(A).$$
Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.
answered yesterday
amrsa
3,4052518
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As Kevin Long wrote in a comment, you already have that $U(A) subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A subseteq B$ implies $U(B) subseteq U(A)$,
$$A subseteq L(U(A)) Longrightarrow U(L(U(A))) subseteq U(A).$$
Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.
answered yesterday
amrsa
3,4052518
As Kevin Long wrote in a comment, you already have that $U(A) subseteq U(L(U(A)))$.
For the converse inclusion, using (1), and your observation that $A subseteq B$ implies $U(B) subseteq U(A)$,
$$A subseteq L(U(A)) Longrightarrow U(L(U(A))) subseteq U(A).$$
Notice also that this is true for any poset, which is quite clear since nowhere in the proof is any lattice property used.
answered yesterday
amrsa
3,4052518
answered yesterday
amrsa
3,4052518
answered yesterday
amrsa
3,4052518
answered yesterday
amrsa
3,4052518
3,4052518
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
add a comment |
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Can you please elaborate a little on the first step? $b in U(A) implies forall a in A: aleq b implies ... ?... implies b in U(L(U(A)))$ I don't fully get what arguments should be applied here.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
Figured it out by myself, thanks.
– Sergey Dylda
yesterday
add a comment |
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1
Your proofs look valid to me. Using what you showed in part 1, you can conclude that $U(A)subseteq U(L(U(A)))$, so only the reverse inclusion is needed. Then you can just ask yourself if you can have $ain U(A)$ such that $anotin U(L(U(A)))$. In other words, can there be $bin L(U(A))$ such that $angeq b$?
– Kevin Long
yesterday
@KevinLong Isn't it backwards from part 1: $A subseteq C(A) implies U(C(A)) subseteq U(A)$ and not the other way around? Because upper/lower bounds of superset are definitely upper/lower bounds of each subset bot not the other way.
– Sergey Dylda
yesterday
Sorry, that's right. I was thinking the right thing, but wrote the wrong one. Everything after that still explains what you need to do.
– Kevin Long
yesterday