Infimum of a union of sets
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Considering the set
$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $
I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.
Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?
real-analysis
edited Nov 17 at 10:39
amWhy
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
|
show 2 more comments
up vote
0
down vote
favorite
Considering the set
$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $
I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.
Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?
real-analysis
edited Nov 17 at 10:39
amWhy
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
3
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
2
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Considering the set
$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $
I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.
Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?
real-analysis
edited Nov 17 at 10:39
amWhy
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
Considering the set
$ A = bigcup_{n=1}^infty (1-frac{1}{n}, n-sqrt n] $
I am trying to prove that $inf(A) = 0$ but for $ epsilon = frac{1}{4}$ it seems there is not an element $x in A$ such that $x$ is smaller than $inf (A) + frac{1}{4}$ but larger than $0$.
Does this mean that the infimum cannot be $0$? Perhaps I am confused by the definition. Thanks for the help. And as a side note, does the set $(0,0]$, which is clopen, contain the point $0$?
real-analysis
real-analysis
edited Nov 17 at 10:39
amWhy
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
edited Nov 17 at 10:39
amWhy
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
edited Nov 17 at 10:39
amWhy
191k27223437
edited Nov 17 at 10:39
amWhy
191k27223437
edited Nov 17 at 10:39
amWhy
191k27223437
191k27223437
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
asked Jan 29 '16 at 3:58
IntegrateThis
1,6971717
1,6971717
When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
3
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
2
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06
|
show 2 more comments
When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
3
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
2
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06
When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
3
3
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
2
2
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06
|
show 2 more comments
1 Answer
1
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0
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Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.
Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.
answered Jan 29 '16 at 4:32
M10687
2,6481720
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.
Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.
answered Jan 29 '16 at 4:32
M10687
2,6481720
add a comment |
up vote
0
down vote
Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.
Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.
answered Jan 29 '16 at 4:32
M10687
2,6481720
add a comment |
up vote
0
down vote
up vote
0
down vote
Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.
Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.
answered Jan 29 '16 at 4:32
M10687
2,6481720
Take $x in ( frac{1}{2}, 2- sqrt{2})$. Then: $x> frac{1}{2}$ $Rightarrow$ $x>frac{x+frac{1}{2}}{2} > frac{1}{2}$. Then for any $x in (frac{1}{2}, 2- sqrt{2})$ $exists$ $y in (frac{1}{2}, 2- sqrt{2})$ such that $x>y> frac{1}{2}$. Then $frac{1}{2} = inf(frac{1}{2}, 2- sqrt{2})$.
Then $inf(A) = frac{1}{2}$, since given $y in A$, $y geq x$ for some $x in (2, 2- sqrt{2})$.
answered Jan 29 '16 at 4:32
M10687
2,6481720
answered Jan 29 '16 at 4:32
M10687
2,6481720
answered Jan 29 '16 at 4:32
M10687
2,6481720
answered Jan 29 '16 at 4:32
M10687
2,6481720
2,6481720
add a comment |
add a comment |
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When $n=1$, you have the set $(0,0]$. What, exactly, is in this set?
– Alex S
Jan 29 '16 at 4:04
3
The set $(0,0]$ should be considered empty.
– OnoL
Jan 29 '16 at 4:05
I'm not sure if clopen sets of that form contain the point or not.
– IntegrateThis
Jan 29 '16 at 4:05
aw ok thus the source of my confusion.
– IntegrateThis
Jan 29 '16 at 4:05
2
When $n=1$, the corresponding set it $(0,0]={xinmathbb{R}:0<xleq 0}$ which is absurd.
– user238841
Jan 29 '16 at 4:06