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If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.
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Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25]
If $A$ is an ordered set has the Least upper bound property iff it has the greatest
lower bound property.
An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$ of $A$ that is bounded above has least upper bound. $A$ satisfies this property. Suppose $B_0$ is an arbitrary set having lower bound. We need to prove it has the greatest lower bound. Suppose $Lsubset A$ is the set of all lower bounds of the set $B_0$. $$exists b_0in B_0, forall xin L, xleq b_0. $$
So, $L$ is bounded above. Hence, It has least upper bound. Let $b$ be the least upper bound of $L$. $$ forall xin L, xleq b. $$ If $exists b'in B_0, b'leq b,$ which contradict the fact that $b$ is the least upper bound of $L$. Hence,$$bleq y, forall yin B_0.$$ So, $b$ is the lower bound. No element of $yin L$ satisfy $ bleq y$. Hence $b$ is the greatest lower bound. If this proof is correct, I can use the similar argument for the converse. I request you to verify my proof.
proof-verification order-theory supremum-and-infimum
asked Jul 21 at 3:26
N. Maneesh
2,4371924
add a comment |Â
up vote
0
down vote
favorite
Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25]
If $A$ is an ordered set has the Least upper bound property iff it has the greatest
lower bound property.
An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$ of $A$ that is bounded above has least upper bound. $A$ satisfies this property. Suppose $B_0$ is an arbitrary set having lower bound. We need to prove it has the greatest lower bound. Suppose $Lsubset A$ is the set of all lower bounds of the set $B_0$. $$exists b_0in B_0, forall xin L, xleq b_0. $$
So, $L$ is bounded above. Hence, It has least upper bound. Let $b$ be the least upper bound of $L$. $$ forall xin L, xleq b. $$ If $exists b'in B_0, b'leq b,$ which contradict the fact that $b$ is the least upper bound of $L$. Hence,$$bleq y, forall yin B_0.$$ So, $b$ is the lower bound. No element of $yin L$ satisfy $ bleq y$. Hence $b$ is the greatest lower bound. If this proof is correct, I can use the similar argument for the converse. I request you to verify my proof.
proof-verification order-theory supremum-and-infimum
asked Jul 21 at 3:26
N. Maneesh
2,4371924
1
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25]
If $A$ is an ordered set has the Least upper bound property iff it has the greatest
lower bound property.
An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$ of $A$ that is bounded above has least upper bound. $A$ satisfies this property. Suppose $B_0$ is an arbitrary set having lower bound. We need to prove it has the greatest lower bound. Suppose $Lsubset A$ is the set of all lower bounds of the set $B_0$. $$exists b_0in B_0, forall xin L, xleq b_0. $$
So, $L$ is bounded above. Hence, It has least upper bound. Let $b$ be the least upper bound of $L$. $$ forall xin L, xleq b. $$ If $exists b'in B_0, b'leq b,$ which contradict the fact that $b$ is the least upper bound of $L$. Hence,$$bleq y, forall yin B_0.$$ So, $b$ is the lower bound. No element of $yin L$ satisfy $ bleq y$. Hence $b$ is the greatest lower bound. If this proof is correct, I can use the similar argument for the converse. I request you to verify my proof.
proof-verification order-theory supremum-and-infimum
asked Jul 21 at 3:26
N. Maneesh
2,4371924
Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25]
If $A$ is an ordered set has the Least upper bound property iff it has the greatest
lower bound property.
An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$ of $A$ that is bounded above has least upper bound. $A$ satisfies this property. Suppose $B_0$ is an arbitrary set having lower bound. We need to prove it has the greatest lower bound. Suppose $Lsubset A$ is the set of all lower bounds of the set $B_0$. $$exists b_0in B_0, forall xin L, xleq b_0. $$
So, $L$ is bounded above. Hence, It has least upper bound. Let $b$ be the least upper bound of $L$. $$ forall xin L, xleq b. $$ If $exists b'in B_0, b'leq b,$ which contradict the fact that $b$ is the least upper bound of $L$. Hence,$$bleq y, forall yin B_0.$$ So, $b$ is the lower bound. No element of $yin L$ satisfy $ bleq y$. Hence $b$ is the greatest lower bound. If this proof is correct, I can use the similar argument for the converse. I request you to verify my proof.
proof-verification order-theory supremum-and-infimum
asked Jul 21 at 3:26
N. Maneesh
2,4371924
edited Jul 21 at 6:10
asked Jul 21 at 3:26
N. Maneesh
2,4371924
asked Jul 21 at 3:26
N. Maneesh
2,4371924
asked Jul 21 at 3:26
N. Maneesh
2,4371924
2,4371924
1
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08
add a comment |Â
1
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08
1
1
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08
add a comment |Â
1 Answer
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The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = x in A: forall b in B: x le b$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 in B$ ($B$ is non-empty), and any $l in L(B)$ we have $l le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = operatornamelub(L(B))$ exists. Claim: $l_0 = operatornameglb(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = operatornameglb(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $operatornamelub(B) = operatornameglb(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = x in A: forall b in B: x le b$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 in B$ ($B$ is non-empty), and any $l in L(B)$ we have $l le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = operatornamelub(L(B))$ exists. Claim: $l_0 = operatornameglb(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = operatornameglb(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $operatornamelub(B) = operatornameglb(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
add a comment |Â
up vote
1
down vote
accepted
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = x in A: forall b in B: x le b$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 in B$ ($B$ is non-empty), and any $l in L(B)$ we have $l le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = operatornamelub(L(B))$ exists. Claim: $l_0 = operatornameglb(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = operatornameglb(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $operatornamelub(B) = operatornameglb(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = x in A: forall b in B: x le b$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 in B$ ($B$ is non-empty), and any $l in L(B)$ we have $l le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = operatornamelub(L(B))$ exists. Claim: $l_0 = operatornameglb(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = operatornameglb(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $operatornamelub(B) = operatornameglb(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
The idea is correct, but could be written down more clearly:
Let $A$ be an ordered set that satisfies the lub property.
Then $A$ satisfies the glb property:
Let $B$ be a non-empty subset of $A$ that is bounded below, define
$$L(B) = x in A: forall b in B: x le b$$ which is the set of lower bounds for $B$, which is by assumption non-empty.
Then for any fixed $b_0 in B$ ($B$ is non-empty), and any $l in L(B)$ we have $l le b_0$. But this says that $L(B)$ is bounded above. So the lub property says that $l_0 = operatornamelub(L(B))$ exists. Claim: $l_0 = operatornameglb(B)$.
For this we need to show two things: $l_0$ is a lower bound for $B$ and there is no greater lower bound for $B$.
So let $b in B$. Then, as before, $b$ as before is an upper bound for $L(B)$ by the definition of $L(B)$, and as $l_0$ is the least upper bound for $L(B)$ we have $l_0 le b$. As $b$ was arbitary, $l_0$ is a lower bound for $B$.
Now suppose $m$ is any lower bound for $B$. Then again by definition, $m in L(B)$.
As $l_0$ is the least upper bound for $L(B)$, it is an upperbound for $L(B)$ so $m le l_0$. So all lower bounds for $B$ are below $l_0$.
Together this says that $l_0 = operatornameglb(B)$ and so $A$ satisfies the glb-property.
The reverse is indeed similar: $operatornamelub(B) = operatornameglb(U(B))$, where $U(B)$ is the set of upperbounds for $B$.
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
answered Jul 21 at 5:15
Henno Brandsma
91.5k342100
91.5k342100
add a comment |Â
add a comment |Â
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1
" Suppose $Lsubset A $ is the lower bound of the set $B_0$". Is that a typo? It doesn't make sense for a set to be a lower bound. Did you mean that $L $ was the set of all lower bounds?
– fleablood
Jul 21 at 5:00
sorry! I meant to say it was set of all lower bounds.
– N. Maneesh
Jul 21 at 6:08