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$E$ is a nonempty subset of an ordered set. If $alpha$ is a lower bound of $E$ and $beta$ is an upper bound, then $alpha leq beta$.
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I think I may be citing the law of transitivity incorrectly in this proof.
Theorem. Let $E$ be a nonempty subset of an ordered set. Suppose $alpha$ is a lower bound of $E$ and $beta$ is an upper bound of $E$. Prove that $alpha leq beta$.
Proof. Since $alpha$ is a lower bound of $E$, and $E$ is nonempty, we have that $forall x in E, x geq alpha$. Similarly, since $beta$ is an upper bound, we have $forall x in E, x leq beta$. Stringing these together yields $forall x in E, alpha leq x leq beta$, which by the transitivity property of ordered sets yields $alpha leq beta$.
Here are my questions on this:
(a) Is it a fair assumption that a subset of an ordered set is also ordered? I wasn't sure whether I ought to 'prove' this first, though it seems almost too trivial to prove.
(b) I use transitivity a bit differently from Rudin's definition, based on past formulations I've seen. He seems to define this as $x < y wedge y < z implies x < z$, i.e., with strict inequalities. Should I break this into cases, wherein I consider all possible cases of strict inequalities as well as equalities between $alpha$, $x$, and $beta$? Or is this a natural extension of this property?
Thanks.
real-analysis proof-explanation supremum-and-infimum
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
asked Aug 18 at 17:06
Matt.P
930414
add a comment |Â
up vote
2
down vote
favorite
I think I may be citing the law of transitivity incorrectly in this proof.
Theorem. Let $E$ be a nonempty subset of an ordered set. Suppose $alpha$ is a lower bound of $E$ and $beta$ is an upper bound of $E$. Prove that $alpha leq beta$.
Proof. Since $alpha$ is a lower bound of $E$, and $E$ is nonempty, we have that $forall x in E, x geq alpha$. Similarly, since $beta$ is an upper bound, we have $forall x in E, x leq beta$. Stringing these together yields $forall x in E, alpha leq x leq beta$, which by the transitivity property of ordered sets yields $alpha leq beta$.
Here are my questions on this:
(a) Is it a fair assumption that a subset of an ordered set is also ordered? I wasn't sure whether I ought to 'prove' this first, though it seems almost too trivial to prove.
(b) I use transitivity a bit differently from Rudin's definition, based on past formulations I've seen. He seems to define this as $x < y wedge y < z implies x < z$, i.e., with strict inequalities. Should I break this into cases, wherein I consider all possible cases of strict inequalities as well as equalities between $alpha$, $x$, and $beta$? Or is this a natural extension of this property?
Thanks.
real-analysis proof-explanation supremum-and-infimum
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
asked Aug 18 at 17:06
Matt.P
930414
It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I think I may be citing the law of transitivity incorrectly in this proof.
Theorem. Let $E$ be a nonempty subset of an ordered set. Suppose $alpha$ is a lower bound of $E$ and $beta$ is an upper bound of $E$. Prove that $alpha leq beta$.
Proof. Since $alpha$ is a lower bound of $E$, and $E$ is nonempty, we have that $forall x in E, x geq alpha$. Similarly, since $beta$ is an upper bound, we have $forall x in E, x leq beta$. Stringing these together yields $forall x in E, alpha leq x leq beta$, which by the transitivity property of ordered sets yields $alpha leq beta$.
Here are my questions on this:
(a) Is it a fair assumption that a subset of an ordered set is also ordered? I wasn't sure whether I ought to 'prove' this first, though it seems almost too trivial to prove.
(b) I use transitivity a bit differently from Rudin's definition, based on past formulations I've seen. He seems to define this as $x < y wedge y < z implies x < z$, i.e., with strict inequalities. Should I break this into cases, wherein I consider all possible cases of strict inequalities as well as equalities between $alpha$, $x$, and $beta$? Or is this a natural extension of this property?
Thanks.
real-analysis proof-explanation supremum-and-infimum
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
asked Aug 18 at 17:06
Matt.P
930414
I think I may be citing the law of transitivity incorrectly in this proof.
Theorem. Let $E$ be a nonempty subset of an ordered set. Suppose $alpha$ is a lower bound of $E$ and $beta$ is an upper bound of $E$. Prove that $alpha leq beta$.
Proof. Since $alpha$ is a lower bound of $E$, and $E$ is nonempty, we have that $forall x in E, x geq alpha$. Similarly, since $beta$ is an upper bound, we have $forall x in E, x leq beta$. Stringing these together yields $forall x in E, alpha leq x leq beta$, which by the transitivity property of ordered sets yields $alpha leq beta$.
Here are my questions on this:
(a) Is it a fair assumption that a subset of an ordered set is also ordered? I wasn't sure whether I ought to 'prove' this first, though it seems almost too trivial to prove.
(b) I use transitivity a bit differently from Rudin's definition, based on past formulations I've seen. He seems to define this as $x < y wedge y < z implies x < z$, i.e., with strict inequalities. Should I break this into cases, wherein I consider all possible cases of strict inequalities as well as equalities between $alpha$, $x$, and $beta$? Or is this a natural extension of this property?
Thanks.
real-analysis proof-explanation supremum-and-infimum
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
asked Aug 18 at 17:06
Matt.P
930414
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
edited Aug 18 at 20:16
Asaf Karagila♦
294k31410737
294k31410737
asked Aug 18 at 17:06
Matt.P
930414
asked Aug 18 at 17:06
Matt.P
930414
asked Aug 18 at 17:06
Matt.P
930414
930414
It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15
add a comment |Â
It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15
It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15
It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $mathbbR$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.
For your second question, to be completely rigorous, the transitivity of $leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
add a comment |Â
up vote
5
down vote
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order
answered Aug 18 at 17:15
Edouardb
480314
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $mathbbR$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.
For your second question, to be completely rigorous, the transitivity of $leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
add a comment |Â
up vote
2
down vote
accepted
For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $mathbbR$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.
For your second question, to be completely rigorous, the transitivity of $leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $mathbbR$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.
For your second question, to be completely rigorous, the transitivity of $leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
For your first question, your ordering actually relies on the ordering on the larger set, since the upper and lower bounds need not be members of the subset $E$ itself! For instance, if the original set is $mathbbR$ and the subset is $(0,1)$, your lower bound could be $0$. So really you rely on the fact that the ordering is defined on the original set.
For your second question, to be completely rigorous, the transitivity of $leq$ follows from transitivity of $<$ and $=$. Personally, I think it is fine to simply leave it as you have, but perhaps your professor wants you to add in additional detail if this is a first course in proofs or something.
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
answered Aug 18 at 17:15
ThomasGrubb
9,78011335
9,78011335
add a comment |Â
add a comment |Â
up vote
5
down vote
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order
answered Aug 18 at 17:15
Edouardb
480314
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
add a comment |Â
up vote
5
down vote
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order
answered Aug 18 at 17:15
Edouardb
480314
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
add a comment |Â
up vote
5
down vote
up vote
5
down vote
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order
answered Aug 18 at 17:15
Edouardb
480314
You're proof is almost correct.
The fact that E is nonempty is useless to state that ∀x∈E, x≥α (it would also be the case if E were empty!). Same for ∀x∈E, x≤β. These two relations are the definition of lower and upper bounds, respectively.
However, you need the fact that E is nonempty to prove that α≤β, because ∀x∈E,α≤x≤β would be also true if E were empty, but you couldn't argue that α≤β (because in this case α≤x≤β would never happen). A correct statement would be "E is nonempty so ∃x∈E. So α≤x and x≤β so α≤β."
(a) Yes, you can prove it easily.
(b) "<" and ">" are not orders because they do not check the reflexivity property of an order
answered Aug 18 at 17:15
Edouardb
480314
edited Aug 20 at 5:47
answered Aug 18 at 17:15
Edouardb
480314
answered Aug 18 at 17:15
Edouardb
480314
answered Aug 18 at 17:15
Edouardb
480314
480314
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
add a comment |Â
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
1
1
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
They can be considered as strict orders. In fact, some books define orders as $<$ instead of $leq$.
– Dog_69
Aug 18 at 17:28
1
1
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
Would you mind explaining why they don't satisfy antisymmetry? It seems that $a < b$, for example, cannot coexist with $b < a$. Surely $<$ wouldn't be reflexive, though, whereas $leq$ would be.
– Matt.P
Aug 18 at 17:33
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
@Matt.P Oh, very good point. I think he/s wanted to say reflexivity.
– Dog_69
Aug 18 at 17:47
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
Yes, my mistake I wanted to say reflexivity. I fixed it, thx
– Edouardb
Aug 20 at 5:48
add a comment |Â
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It seems good. Regarding your second question, it is a natural extension.
– Dog_69
Aug 18 at 17:15