Let $A$ be a set and $s$ its supremum. Given an $epsilon>0$, can I assure that I can find and $ain A$ that...
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Let $A$ be a set and $s$ it's supreme. Given an $epsilon>0$, can I assure that I can find and $ain A$ that meets: $s-epsilon<a<s$ ?
I tried saying that $s>a,, forall ain A$, and also, seeing that we don't know if the set A is lower bounded if I fix and $a$ I'll always be able to pick and $epsilon$ big enough to mate it true, but... that's enough?
I believe I need something more to demonstrate (or negate) it, any help?
real-analysis upper-lower-bounds
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
asked Nov 13 at 16:19
iggykimi
726
add a comment |
up vote
0
down vote
favorite
Let $A$ be a set and $s$ it's supreme. Given an $epsilon>0$, can I assure that I can find and $ain A$ that meets: $s-epsilon<a<s$ ?
I tried saying that $s>a,, forall ain A$, and also, seeing that we don't know if the set A is lower bounded if I fix and $a$ I'll always be able to pick and $epsilon$ big enough to mate it true, but... that's enough?
I believe I need something more to demonstrate (or negate) it, any help?
real-analysis upper-lower-bounds
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
asked Nov 13 at 16:19
iggykimi
726
A set of what?
– Bernard
Nov 13 at 16:24
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
1
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ be a set and $s$ it's supreme. Given an $epsilon>0$, can I assure that I can find and $ain A$ that meets: $s-epsilon<a<s$ ?
I tried saying that $s>a,, forall ain A$, and also, seeing that we don't know if the set A is lower bounded if I fix and $a$ I'll always be able to pick and $epsilon$ big enough to mate it true, but... that's enough?
I believe I need something more to demonstrate (or negate) it, any help?
real-analysis upper-lower-bounds
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
asked Nov 13 at 16:19
iggykimi
726
Let $A$ be a set and $s$ it's supreme. Given an $epsilon>0$, can I assure that I can find and $ain A$ that meets: $s-epsilon<a<s$ ?
I tried saying that $s>a,, forall ain A$, and also, seeing that we don't know if the set A is lower bounded if I fix and $a$ I'll always be able to pick and $epsilon$ big enough to mate it true, but... that's enough?
I believe I need something more to demonstrate (or negate) it, any help?
real-analysis upper-lower-bounds
real-analysis upper-lower-bounds
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
asked Nov 13 at 16:19
iggykimi
726
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
asked Nov 13 at 16:19
iggykimi
726
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
edited Nov 13 at 16:26
Asaf Karagila♦
299k32420750
299k32420750
asked Nov 13 at 16:19
iggykimi
726
asked Nov 13 at 16:19
iggykimi
726
asked Nov 13 at 16:19
iggykimi
726
726
A set of what?
– Bernard
Nov 13 at 16:24
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
1
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33
add a comment |
A set of what?
– Bernard
Nov 13 at 16:24
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
1
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33
A set of what?
– Bernard
Nov 13 at 16:24
A set of what?
– Bernard
Nov 13 at 16:24
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
1
1
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
No, actually. Let for example $A=[0,1]cup{2}$. Then for $epsilon=tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
add a comment |
up vote
0
down vote
No.
Let $A={42}$. Then $s=sup A = 42$ and for $epsilon:=1$, there is no $ain A$ with $s-epsilon<a<s$.
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No, actually. Let for example $A=[0,1]cup{2}$. Then for $epsilon=tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
add a comment |
up vote
2
down vote
accepted
No, actually. Let for example $A=[0,1]cup{2}$. Then for $epsilon=tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No, actually. Let for example $A=[0,1]cup{2}$. Then for $epsilon=tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
No, actually. Let for example $A=[0,1]cup{2}$. Then for $epsilon=tfrac12$, such element $a$ does not exist. If there is a sequence of elements in $A$ converging to $s$, then you could prove the statement you conjectured.
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
answered Nov 13 at 16:26
vrugtehagel
10.7k1548
10.7k1548
add a comment |
add a comment |
up vote
0
down vote
No.
Let $A={42}$. Then $s=sup A = 42$ and for $epsilon:=1$, there is no $ain A$ with $s-epsilon<a<s$.
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
add a comment |
up vote
0
down vote
No.
Let $A={42}$. Then $s=sup A = 42$ and for $epsilon:=1$, there is no $ain A$ with $s-epsilon<a<s$.
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
add a comment |
up vote
0
down vote
up vote
0
down vote
No.
Let $A={42}$. Then $s=sup A = 42$ and for $epsilon:=1$, there is no $ain A$ with $s-epsilon<a<s$.
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
No.
Let $A={42}$. Then $s=sup A = 42$ and for $epsilon:=1$, there is no $ain A$ with $s-epsilon<a<s$.
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
answered Nov 13 at 16:26
Hagen von Eitzen
273k21266493
273k21266493
add a comment |
add a comment |
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A set of what?
– Bernard
Nov 13 at 16:24
Of real numbers, I suppose.
– vrugtehagel
Nov 13 at 16:27
1
You can only find $s-epsilon<ale s$.
– Eclipse Sun
Nov 13 at 16:30
I guess you can find your $a$ if $snotin A$
– krirkrirk
Nov 13 at 16:33