Mixing
Real Analysis Convergance
$begingroup$
If E is a subset of R, nonempty, and bounded, then there exists e in E such that supE-1 < e < supE.
I understand how this works for the subset with natural numbers because for the less than or equal to inequality, supE-1= e. However, I do not understand how this can work for all subsets of R where the inequality is strictly less than.
I was thinking that this possibly had something to do with the Archimedean principle that for every x in R there is an n in N such that n>x. So there must be a real number between supE-1 and supE.
Any help is appreciated in advance!
real-analysis supremum-and-infimum
asked Feb 1 at 1:49
lj_growllj_growl
607
$endgroup$
add a comment |
$begingroup$
If E is a subset of R, nonempty, and bounded, then there exists e in E such that supE-1 < e < supE.
I understand how this works for the subset with natural numbers because for the less than or equal to inequality, supE-1= e. However, I do not understand how this can work for all subsets of R where the inequality is strictly less than.
I was thinking that this possibly had something to do with the Archimedean principle that for every x in R there is an n in N such that n>x. So there must be a real number between supE-1 and supE.
Any help is appreciated in advance!
real-analysis supremum-and-infimum
asked Feb 1 at 1:49
lj_growllj_growl
607
$endgroup$
1
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
2
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
1
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04
add a comment |
$begingroup$
If E is a subset of R, nonempty, and bounded, then there exists e in E such that supE-1 < e < supE.
I understand how this works for the subset with natural numbers because for the less than or equal to inequality, supE-1= e. However, I do not understand how this can work for all subsets of R where the inequality is strictly less than.
I was thinking that this possibly had something to do with the Archimedean principle that for every x in R there is an n in N such that n>x. So there must be a real number between supE-1 and supE.
Any help is appreciated in advance!
real-analysis supremum-and-infimum
asked Feb 1 at 1:49
lj_growllj_growl
607
$endgroup$
If E is a subset of R, nonempty, and bounded, then there exists e in E such that supE-1 < e < supE.
I understand how this works for the subset with natural numbers because for the less than or equal to inequality, supE-1= e. However, I do not understand how this can work for all subsets of R where the inequality is strictly less than.
I was thinking that this possibly had something to do with the Archimedean principle that for every x in R there is an n in N such that n>x. So there must be a real number between supE-1 and supE.
Any help is appreciated in advance!
real-analysis supremum-and-infimum
real-analysis supremum-and-infimum
asked Feb 1 at 1:49
lj_growllj_growl
607
asked Feb 1 at 1:49
lj_growllj_growl
607
asked Feb 1 at 1:49
lj_growllj_growl
607
asked Feb 1 at 1:49
lj_growllj_growl
607
asked Feb 1 at 1:49
lj_growllj_growl
607
607
1
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
2
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
1
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04
add a comment |
1
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
2
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
1
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04
1
1
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
2
2
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
1
1
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04
add a comment |
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1
$begingroup$
Hint: if no $e in E$ existed such that $sup E - 1 < e le sup E$, then $sup E - 1$ is an upper bound for $E$...
$endgroup$
– Theo Bendit
Feb 1 at 1:51
$begingroup$
@TheoBendit Why did I not think of that lol. But question: is there a difference because all the inequalities are strictly less than opposed to having one with the less than/equal to? It may just be a typo on the question, but I don't see how there would have to be an e if this were the natural numbers or integers. Either supE -1=e or e=supE for those sets right?
$endgroup$
– lj_growl
Feb 1 at 2:00
2
$begingroup$
If you don't have $le$, and just $<$, then the result is false! For example, take a singleton set, e.g. ${0}$. Or, perhaps, take the set $[0, 1] cup {3}$.
$endgroup$
– Theo Bendit
Feb 1 at 2:03
1
$begingroup$
@TheoBendit That makes sense. I believe that the question has a typo. Thank you so much for your help!
$endgroup$
– lj_growl
Feb 1 at 2:04