Real Analysis: Supremum Limit Points
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Let B = $dfrac{n-1}{n} , n in mathbb N$.
Consider A = $(0,1)$ B.
Find the supremum, infimum, and limit points of A. Is A open, closed, or neither?
Solution Attempt: Sup $= 1$, inf $= 0$, The limit points are $1$ and $0$. The set is open.
real-analysis
edited Nov 18 at 3:36
Is12Prime
12619
asked Nov 8 at 21:10
XiangLiang
575
add a comment |
up vote
1
down vote
favorite
Let B = $dfrac{n-1}{n} , n in mathbb N$.
Consider A = $(0,1)$ B.
Find the supremum, infimum, and limit points of A. Is A open, closed, or neither?
Solution Attempt: Sup $= 1$, inf $= 0$, The limit points are $1$ and $0$. The set is open.
real-analysis
edited Nov 18 at 3:36
Is12Prime
12619
asked Nov 8 at 21:10
XiangLiang
575
Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let B = $dfrac{n-1}{n} , n in mathbb N$.
Consider A = $(0,1)$ B.
Find the supremum, infimum, and limit points of A. Is A open, closed, or neither?
Solution Attempt: Sup $= 1$, inf $= 0$, The limit points are $1$ and $0$. The set is open.
real-analysis
edited Nov 18 at 3:36
Is12Prime
12619
asked Nov 8 at 21:10
XiangLiang
575
Let B = $dfrac{n-1}{n} , n in mathbb N$.
Consider A = $(0,1)$ B.
Find the supremum, infimum, and limit points of A. Is A open, closed, or neither?
Solution Attempt: Sup $= 1$, inf $= 0$, The limit points are $1$ and $0$. The set is open.
real-analysis
real-analysis
edited Nov 18 at 3:36
Is12Prime
12619
asked Nov 8 at 21:10
XiangLiang
575
edited Nov 18 at 3:36
Is12Prime
12619
asked Nov 8 at 21:10
XiangLiang
575
edited Nov 18 at 3:36
Is12Prime
12619
edited Nov 18 at 3:36
Is12Prime
12619
edited Nov 18 at 3:36
Is12Prime
12619
12619
asked Nov 8 at 21:10
XiangLiang
575
asked Nov 8 at 21:10
XiangLiang
575
asked Nov 8 at 21:10
XiangLiang
575
575
Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49
add a comment |
Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49
Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49
add a comment |
1 Answer
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up vote
1
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Your answers for supremum and infimum of $A$ are correct. The answer for limit points is also correct, assuming you are using the following definition:
$x in mathbb{R}$ is a limit point of $A subseteq mathbb{R}$ iff for every $r > 0$, the open interval $(x - r, x + r)$ intersects $A$ at a point other than $x$.
From this definition it should follow that every point interior to a set is also a limit point of that set. Since $A$ is an open set every $x in A$ is interior to $A$, so every $x in A$ is a limit point of $A$.
Other than $A$ itself, all elements of $B cup {0, 1}$ are also limit points. So $[0, 1]$ is the set of all limit points of $A$.
answered Nov 18 at 3:11
Akhil Jalan
737
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your answers for supremum and infimum of $A$ are correct. The answer for limit points is also correct, assuming you are using the following definition:
$x in mathbb{R}$ is a limit point of $A subseteq mathbb{R}$ iff for every $r > 0$, the open interval $(x - r, x + r)$ intersects $A$ at a point other than $x$.
From this definition it should follow that every point interior to a set is also a limit point of that set. Since $A$ is an open set every $x in A$ is interior to $A$, so every $x in A$ is a limit point of $A$.
Other than $A$ itself, all elements of $B cup {0, 1}$ are also limit points. So $[0, 1]$ is the set of all limit points of $A$.
answered Nov 18 at 3:11
Akhil Jalan
737
add a comment |
up vote
1
down vote
Your answers for supremum and infimum of $A$ are correct. The answer for limit points is also correct, assuming you are using the following definition:
$x in mathbb{R}$ is a limit point of $A subseteq mathbb{R}$ iff for every $r > 0$, the open interval $(x - r, x + r)$ intersects $A$ at a point other than $x$.
From this definition it should follow that every point interior to a set is also a limit point of that set. Since $A$ is an open set every $x in A$ is interior to $A$, so every $x in A$ is a limit point of $A$.
Other than $A$ itself, all elements of $B cup {0, 1}$ are also limit points. So $[0, 1]$ is the set of all limit points of $A$.
answered Nov 18 at 3:11
Akhil Jalan
737
add a comment |
up vote
1
down vote
up vote
1
down vote
Your answers for supremum and infimum of $A$ are correct. The answer for limit points is also correct, assuming you are using the following definition:
$x in mathbb{R}$ is a limit point of $A subseteq mathbb{R}$ iff for every $r > 0$, the open interval $(x - r, x + r)$ intersects $A$ at a point other than $x$.
From this definition it should follow that every point interior to a set is also a limit point of that set. Since $A$ is an open set every $x in A$ is interior to $A$, so every $x in A$ is a limit point of $A$.
Other than $A$ itself, all elements of $B cup {0, 1}$ are also limit points. So $[0, 1]$ is the set of all limit points of $A$.
answered Nov 18 at 3:11
Akhil Jalan
737
Your answers for supremum and infimum of $A$ are correct. The answer for limit points is also correct, assuming you are using the following definition:
$x in mathbb{R}$ is a limit point of $A subseteq mathbb{R}$ iff for every $r > 0$, the open interval $(x - r, x + r)$ intersects $A$ at a point other than $x$.
From this definition it should follow that every point interior to a set is also a limit point of that set. Since $A$ is an open set every $x in A$ is interior to $A$, so every $x in A$ is a limit point of $A$.
Other than $A$ itself, all elements of $B cup {0, 1}$ are also limit points. So $[0, 1]$ is the set of all limit points of $A$.
answered Nov 18 at 3:11
Akhil Jalan
737
edited Nov 18 at 3:22
answered Nov 18 at 3:11
Akhil Jalan
737
answered Nov 18 at 3:11
Akhil Jalan
737
answered Nov 18 at 3:11
Akhil Jalan
737
737
add a comment |
add a comment |
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Is $B$ the set of all points $frac{n-1}{n}$ with $ninmathbb N$?
– Dave
Nov 8 at 21:14
Your answer for limit points is incorrect. Any $bin B$ should be a limit point of $A$.
– kcborys
Nov 8 at 21:15
Yep, that's right, what is your question?
– rldias
Nov 8 at 21:16
Can someone else confirm if my answer for limit points is correct or not?
– XiangLiang
Nov 8 at 21:38
@Dave. Naw, it looks like a sequence.
– William Elliot
Nov 9 at 8:49