Clarification for the least upper bound property
$begingroup$
I've just started on Principles of Mathematical Analysis by Walter Rudin (Third Edition) and came across the following definition for the least-upper-bound property:
Definition: An ordered set $S$ is said to have the least-upper-bound
property if the following is true:
If $Esubset S$, $E$ not empty, and $E$ is bounded above, then sup $E$
exists in $S$.
My question is the following: Does this mean that open bounded sets of the form $(a,b)$ do not have the least-upper-bound property, since the supremum of the subset $(frac{a+b}{2},b)$ is $bnotin (a,b)$ ?
real-analysis
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
$endgroup$
add a comment |
$begingroup$
I've just started on Principles of Mathematical Analysis by Walter Rudin (Third Edition) and came across the following definition for the least-upper-bound property:
Definition: An ordered set $S$ is said to have the least-upper-bound
property if the following is true:
If $Esubset S$, $E$ not empty, and $E$ is bounded above, then sup $E$
exists in $S$.
My question is the following: Does this mean that open bounded sets of the form $(a,b)$ do not have the least-upper-bound property, since the supremum of the subset $(frac{a+b}{2},b)$ is $bnotin (a,b)$ ?
real-analysis
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
$endgroup$
add a comment |
$begingroup$
I've just started on Principles of Mathematical Analysis by Walter Rudin (Third Edition) and came across the following definition for the least-upper-bound property:
Definition: An ordered set $S$ is said to have the least-upper-bound
property if the following is true:
If $Esubset S$, $E$ not empty, and $E$ is bounded above, then sup $E$
exists in $S$.
My question is the following: Does this mean that open bounded sets of the form $(a,b)$ do not have the least-upper-bound property, since the supremum of the subset $(frac{a+b}{2},b)$ is $bnotin (a,b)$ ?
real-analysis
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
$endgroup$
I've just started on Principles of Mathematical Analysis by Walter Rudin (Third Edition) and came across the following definition for the least-upper-bound property:
Definition: An ordered set $S$ is said to have the least-upper-bound
property if the following is true:
If $Esubset S$, $E$ not empty, and $E$ is bounded above, then sup $E$
exists in $S$.
My question is the following: Does this mean that open bounded sets of the form $(a,b)$ do not have the least-upper-bound property, since the supremum of the subset $(frac{a+b}{2},b)$ is $bnotin (a,b)$ ?
real-analysis
real-analysis
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
edited Jan 7 at 2:32
David C. Ullrich
61.7k44095
61.7k44095
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
asked Jan 6 at 17:36
Sean LeeSean Lee
801214
801214
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $mathbb R$, but this doesn't matter).
In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $Esubseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.
For a more dramatic example, consider $S=(0,1)cup{2}cup(3,4)$. This set also has the least upper bound property. For instance, $sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
$endgroup$
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
add a comment |
$begingroup$
Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!
Look back at the definition: If $Esubset (a,b)=S$ is bounded above in $S$ there exists $cin S$ such that $xle c$ for every $xin E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $mathbb R$, but this doesn't matter).
In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $Esubseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.
For a more dramatic example, consider $S=(0,1)cup{2}cup(3,4)$. This set also has the least upper bound property. For instance, $sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
$endgroup$
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
add a comment |
$begingroup$
The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $mathbb R$, but this doesn't matter).
In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $Esubseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.
For a more dramatic example, consider $S=(0,1)cup{2}cup(3,4)$. This set also has the least upper bound property. For instance, $sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
$endgroup$
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
add a comment |
$begingroup$
The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $mathbb R$, but this doesn't matter).
In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $Esubseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.
For a more dramatic example, consider $S=(0,1)cup{2}cup(3,4)$. This set also has the least upper bound property. For instance, $sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
$endgroup$
The property is defined relative to the set $S$, not to any potential superset of $S$. The example you give has $S=(a,b)$ and the subset you chose as $E$, namely $((a+b)/2,b)$ is not bounded above in $S$ (it is bounded above in $mathbb R$, but this doesn't matter).
In fact, you can easily check that $S=(a,b)$ has the least upper bound property, the point being that if $Esubseteq S$ is nonempty and bounded above (in $S$), then its supremum exists in $mathbb R$, and therefore in $S$, since this supremum is larger than $a$ and strictly smaller than $b$.
For a more dramatic example, consider $S=(0,1)cup{2}cup(3,4)$. This set also has the least upper bound property. For instance, $sup(0,1)=2$. Of course, once we consider $S$ as a subset of the reals, this changes but, again, the property is defined as something intrinsic to $S$, independent of what happens in any potential larger sets containing $S$.
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
answered Jan 6 at 17:53
Andrés E. CaicedoAndrés E. Caicedo
65.9k8160252
65.9k8160252
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
add a comment |
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
$begingroup$
Ah I see; So if I may clarify further, whenever we say a set $E$ is 'bounded above', it is understood to be 'bounded above' relative to some set $S$ where $Esubset S$?
$endgroup$
– Sean Lee
Jan 6 at 18:12
1
1
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
$begingroup$
Yes, being bounded above is relative to whatever the "ambient" universe $S$ is being considered.
$endgroup$
– Andrés E. Caicedo
Jan 6 at 18:13
add a comment |
$begingroup$
Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!
Look back at the definition: If $Esubset (a,b)=S$ is bounded above in $S$ there exists $cin S$ such that $xle c$ for every $xin E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
$begingroup$
Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!
Look back at the definition: If $Esubset (a,b)=S$ is bounded above in $S$ there exists $cin S$ such that $xle c$ for every $xin E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
add a comment |
$begingroup$
Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!
Look back at the definition: If $Esubset (a,b)=S$ is bounded above in $S$ there exists $cin S$ such that $xle c$ for every $xin E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
$endgroup$
Assuming you mean $S=(a,b)$: Actually yes, $(a,b)$ does have the least upper bound property. True, if $E=((a+b)/2,b)$ then there is no $sup E$ in $S$. But that doesn't matter, because (relative to $S$) the set $E$ is not bounded above!
Look back at the definition: If $Esubset (a,b)=S$ is bounded above in $S$ there exists $cin S$ such that $xle c$ for every $xin E$. Saying that $b$ is an upper bound for $((a+b)/2,b)$ doesn't matter, since we're only talking about elements of $S$.
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
answered Jan 6 at 18:06
David C. UllrichDavid C. Ullrich
61.7k44095
61.7k44095
add a comment |
add a comment |
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