Visual interpretation of $sup(A) leq sup(B) $
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By definition for Sup(S):
$forall x in S, xleq sup(S)$ and $forall s_1in S<sup(S), exists x > s_1 $
Is the below visualization of $sup(A)leq sup(B)$ correct? With $X_A in A, X_B in B$.
If so would the following statements about it be equivalent to $Sup(A)leq Sup(B)$ I think they are from the diagram.
- $forall X_B exists X_A : X_B geq X_A $
- $forall X_B $ and $ epsilon >0, exists X_A: X_Aleq X_B + epsilon $
analysis proof-verification
edited Aug 30 at 9:13
Bernard
112k635102
asked Aug 30 at 8:45
glockm15
1649
add a comment |Â
up vote
0
down vote
favorite
By definition for Sup(S):
$forall x in S, xleq sup(S)$ and $forall s_1in S<sup(S), exists x > s_1 $
Is the below visualization of $sup(A)leq sup(B)$ correct? With $X_A in A, X_B in B$.
If so would the following statements about it be equivalent to $Sup(A)leq Sup(B)$ I think they are from the diagram.
- $forall X_B exists X_A : X_B geq X_A $
- $forall X_B $ and $ epsilon >0, exists X_A: X_Aleq X_B + epsilon $
analysis proof-verification
edited Aug 30 at 9:13
Bernard
112k635102
asked Aug 30 at 8:45
glockm15
1649
1
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
1
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
By definition for Sup(S):
$forall x in S, xleq sup(S)$ and $forall s_1in S<sup(S), exists x > s_1 $
Is the below visualization of $sup(A)leq sup(B)$ correct? With $X_A in A, X_B in B$.
If so would the following statements about it be equivalent to $Sup(A)leq Sup(B)$ I think they are from the diagram.
- $forall X_B exists X_A : X_B geq X_A $
- $forall X_B $ and $ epsilon >0, exists X_A: X_Aleq X_B + epsilon $
analysis proof-verification
edited Aug 30 at 9:13
Bernard
112k635102
asked Aug 30 at 8:45
glockm15
1649
By definition for Sup(S):
$forall x in S, xleq sup(S)$ and $forall s_1in S<sup(S), exists x > s_1 $
Is the below visualization of $sup(A)leq sup(B)$ correct? With $X_A in A, X_B in B$.
If so would the following statements about it be equivalent to $Sup(A)leq Sup(B)$ I think they are from the diagram.
- $forall X_B exists X_A : X_B geq X_A $
- $forall X_B $ and $ epsilon >0, exists X_A: X_Aleq X_B + epsilon $
analysis proof-verification
analysis proof-verification
edited Aug 30 at 9:13
Bernard
112k635102
asked Aug 30 at 8:45
glockm15
1649
edited Aug 30 at 9:13
Bernard
112k635102
asked Aug 30 at 8:45
glockm15
1649
edited Aug 30 at 9:13
Bernard
112k635102
edited Aug 30 at 9:13
Bernard
112k635102
edited Aug 30 at 9:13
Bernard
112k635102
112k635102
asked Aug 30 at 8:45
glockm15
1649
asked Aug 30 at 8:45
glockm15
1649
asked Aug 30 at 8:45
glockm15
1649
1649
1
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
1
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24
add a comment |Â
1
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
1
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24
1
1
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
1
1
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Here's a very interesting mistake you are making :
Even $sup A leq sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.
For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $sup A leq sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 in A$.
Lemma : $sup A leq sup B$ implies that for all $a in A, colorbluea neq sup A$, there exists $b in B$ such that $a < b$.
Proof : If $sup A leq sup B$, then fix $a in A$. We know that $a colorred< sup A $, so $a < sup B$ and hence there is an element of $B$ which is greater than $a$.
Lemma : However, $sup A leq sup B$ is implied by : for all $a in A$ there exists $b in B$ such that $a < b$.
Proof : For all $epsilon > 0$, there exists $a > sup A - epsilon$, now find $b > a > sup A - epsilon$, and it follows that $sup B geq b > a > sup A - epsilon$, for all $epsilon$, so the lemma follows.
In conclusion, $sup A leq sup B$ is sandwiched between two very close looking but certainly different conditions :
It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.
It is weaker than the condition that every element of $A$ is to the left of some element of $B$.
Note that if $A$ does not contain its supremum, then the condition $sup A leq sup B$ is equivalent to any of the above conditions.
Therefore, a diagram wishing to illustrate $sup A leq sup B$ must illustrate that it is sandwiched between the two cases.
Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
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
Here's a very interesting mistake you are making :
Even $sup A leq sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.
For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $sup A leq sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 in A$.
Lemma : $sup A leq sup B$ implies that for all $a in A, colorbluea neq sup A$, there exists $b in B$ such that $a < b$.
Proof : If $sup A leq sup B$, then fix $a in A$. We know that $a colorred< sup A $, so $a < sup B$ and hence there is an element of $B$ which is greater than $a$.
Lemma : However, $sup A leq sup B$ is implied by : for all $a in A$ there exists $b in B$ such that $a < b$.
Proof : For all $epsilon > 0$, there exists $a > sup A - epsilon$, now find $b > a > sup A - epsilon$, and it follows that $sup B geq b > a > sup A - epsilon$, for all $epsilon$, so the lemma follows.
In conclusion, $sup A leq sup B$ is sandwiched between two very close looking but certainly different conditions :
It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.
It is weaker than the condition that every element of $A$ is to the left of some element of $B$.
Note that if $A$ does not contain its supremum, then the condition $sup A leq sup B$ is equivalent to any of the above conditions.
Therefore, a diagram wishing to illustrate $sup A leq sup B$ must illustrate that it is sandwiched between the two cases.
Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
add a comment |Â
up vote
1
down vote
accepted
Here's a very interesting mistake you are making :
Even $sup A leq sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.
For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $sup A leq sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 in A$.
Lemma : $sup A leq sup B$ implies that for all $a in A, colorbluea neq sup A$, there exists $b in B$ such that $a < b$.
Proof : If $sup A leq sup B$, then fix $a in A$. We know that $a colorred< sup A $, so $a < sup B$ and hence there is an element of $B$ which is greater than $a$.
Lemma : However, $sup A leq sup B$ is implied by : for all $a in A$ there exists $b in B$ such that $a < b$.
Proof : For all $epsilon > 0$, there exists $a > sup A - epsilon$, now find $b > a > sup A - epsilon$, and it follows that $sup B geq b > a > sup A - epsilon$, for all $epsilon$, so the lemma follows.
In conclusion, $sup A leq sup B$ is sandwiched between two very close looking but certainly different conditions :
It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.
It is weaker than the condition that every element of $A$ is to the left of some element of $B$.
Note that if $A$ does not contain its supremum, then the condition $sup A leq sup B$ is equivalent to any of the above conditions.
Therefore, a diagram wishing to illustrate $sup A leq sup B$ must illustrate that it is sandwiched between the two cases.
Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's a very interesting mistake you are making :
Even $sup A leq sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.
For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $sup A leq sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 in A$.
Lemma : $sup A leq sup B$ implies that for all $a in A, colorbluea neq sup A$, there exists $b in B$ such that $a < b$.
Proof : If $sup A leq sup B$, then fix $a in A$. We know that $a colorred< sup A $, so $a < sup B$ and hence there is an element of $B$ which is greater than $a$.
Lemma : However, $sup A leq sup B$ is implied by : for all $a in A$ there exists $b in B$ such that $a < b$.
Proof : For all $epsilon > 0$, there exists $a > sup A - epsilon$, now find $b > a > sup A - epsilon$, and it follows that $sup B geq b > a > sup A - epsilon$, for all $epsilon$, so the lemma follows.
In conclusion, $sup A leq sup B$ is sandwiched between two very close looking but certainly different conditions :
It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.
It is weaker than the condition that every element of $A$ is to the left of some element of $B$.
Note that if $A$ does not contain its supremum, then the condition $sup A leq sup B$ is equivalent to any of the above conditions.
Therefore, a diagram wishing to illustrate $sup A leq sup B$ must illustrate that it is sandwiched between the two cases.
Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
Here's a very interesting mistake you are making :
Even $sup A leq sup B$ does not guarantee that every element of $A$ lies to the left of some element of $B$.
For example, simply take $A = [0,1]$ and $B = [0,1)$. Then, $sup A leq sup B$ , since both are equal to $1$, but $A$ still contains an element which is bigger than every element of $B$, namely $1 in A$.
Lemma : $sup A leq sup B$ implies that for all $a in A, colorbluea neq sup A$, there exists $b in B$ such that $a < b$.
Proof : If $sup A leq sup B$, then fix $a in A$. We know that $a colorred< sup A $, so $a < sup B$ and hence there is an element of $B$ which is greater than $a$.
Lemma : However, $sup A leq sup B$ is implied by : for all $a in A$ there exists $b in B$ such that $a < b$.
Proof : For all $epsilon > 0$, there exists $a > sup A - epsilon$, now find $b > a > sup A - epsilon$, and it follows that $sup B geq b > a > sup A - epsilon$, for all $epsilon$, so the lemma follows.
In conclusion, $sup A leq sup B$ is sandwiched between two very close looking but certainly different conditions :
It is stronger than the condition that every element of $A$ bar the supremum(if it is in $A$) is strictly to the left of some element of $B$.
It is weaker than the condition that every element of $A$ is to the left of some element of $B$.
Note that if $A$ does not contain its supremum, then the condition $sup A leq sup B$ is equivalent to any of the above conditions.
Therefore, a diagram wishing to illustrate $sup A leq sup B$ must illustrate that it is sandwiched between the two cases.
Therefore, one diagram does not do the job : many are required. While you have demonstrated a single and correct example, it is still worth noting from above that the condition given to you is a very subtle one. I leave you to decide how to draw a diagram after this discussion.
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
answered Aug 30 at 11:01
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.5k32870
33.5k32870
add a comment |Â
add a comment |Â
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1
No. Even though $Sup(B)ge Sup(A)$, $B$ may have elements smaller than every element of $A$. Consider $A=5$ and $B=1,2,3,4,5,6$.
– John Douma
Aug 30 at 8:54
@JohnDouma , oh i see, so in the diagram, assuming that it is correct. we can see that $X_B$ over laps onto Sup(A) and there are no $X_A$ smaller to choose from. So is the diagram correct?
– glockm15
Aug 30 at 9:00
1
I am not sure what constitutes a correct diagram. It does seem to suggest that the greatest lower bound of $B$ is also greater than the greatest lower bound of $A$ which, as the above example shows, is not necessarily true.
– John Douma
Aug 30 at 9:24