What is the difference between “closed ” and “bounded” in terms of domains?
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I'm working on understanding double & triple integrals (and 3-space geometry in general) and I often encounter in my textbooks the requirement that f is continuous on "a closed bounded domain". Wouldn't these concepts be the same thing? Like, if a domain is closed, it contains it's endpoints, and it thus necessarily finite, and if it is bounded it is contained within some "ball" of finite radius centered around the origin and is so finite. I can't really imagine a domain being closed, and not bound, or vice versa. Am I missing some detail in distinguishing these two?
calculus definite-integrals
asked Mar 15 '15 at 10:58
Laplacinator
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up vote
1
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I'm working on understanding double & triple integrals (and 3-space geometry in general) and I often encounter in my textbooks the requirement that f is continuous on "a closed bounded domain". Wouldn't these concepts be the same thing? Like, if a domain is closed, it contains it's endpoints, and it thus necessarily finite, and if it is bounded it is contained within some "ball" of finite radius centered around the origin and is so finite. I can't really imagine a domain being closed, and not bound, or vice versa. Am I missing some detail in distinguishing these two?
calculus definite-integrals
asked Mar 15 '15 at 10:58
Laplacinator
470513
1
$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm working on understanding double & triple integrals (and 3-space geometry in general) and I often encounter in my textbooks the requirement that f is continuous on "a closed bounded domain". Wouldn't these concepts be the same thing? Like, if a domain is closed, it contains it's endpoints, and it thus necessarily finite, and if it is bounded it is contained within some "ball" of finite radius centered around the origin and is so finite. I can't really imagine a domain being closed, and not bound, or vice versa. Am I missing some detail in distinguishing these two?
calculus definite-integrals
asked Mar 15 '15 at 10:58
Laplacinator
470513
I'm working on understanding double & triple integrals (and 3-space geometry in general) and I often encounter in my textbooks the requirement that f is continuous on "a closed bounded domain". Wouldn't these concepts be the same thing? Like, if a domain is closed, it contains it's endpoints, and it thus necessarily finite, and if it is bounded it is contained within some "ball" of finite radius centered around the origin and is so finite. I can't really imagine a domain being closed, and not bound, or vice versa. Am I missing some detail in distinguishing these two?
calculus definite-integrals
calculus definite-integrals
asked Mar 15 '15 at 10:58
Laplacinator
470513
asked Mar 15 '15 at 10:58
Laplacinator
470513
asked Mar 15 '15 at 10:58
Laplacinator
470513
asked Mar 15 '15 at 10:58
Laplacinator
470513
asked Mar 15 '15 at 10:58
Laplacinator
470513
470513
1
$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40
|
show 2 more comments
1
$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40
1
1
$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40
|
show 2 more comments
1 Answer
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In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
answered Oct 22 at 13:39
Arpan Ganguli
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In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
answered Oct 22 at 13:39
Arpan Ganguli
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up vote
0
down vote
In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
answered Oct 22 at 13:39
Arpan Ganguli
113
add a comment |
up vote
0
down vote
up vote
0
down vote
In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
answered Oct 22 at 13:39
Arpan Ganguli
113
In a bounded set, the endpoints need not necessarily be a part of the set whereas in a closed set, the endpoints need to be a part of that set (as you have mentioned in your question). E.g. [0,1] and [0,1) are both bounded (by 0 and 1), but the second set isn't closed.
answered Oct 22 at 13:39
Arpan Ganguli
113
answered Oct 22 at 13:39
Arpan Ganguli
113
answered Oct 22 at 13:39
Arpan Ganguli
113
answered Oct 22 at 13:39
Arpan Ganguli
113
113
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add a comment |
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$mathbb{R}$ is closed, but not bounded
– qwenty
Mar 15 '15 at 11:01
Furthermore, $(0,1)$ is bounded, but not closed.
– Relure
Mar 15 '15 at 11:04
In $mathbf R^n$, a closed bounded domain is exactly a compact domain, it it has many wonderful properties: a numerical function has a maximum and a minimum, every sequence has a convergent subsequence, &c.
– Bernard
Mar 15 '15 at 11:10
The plane ${y = x}$, the sphere ${|mathbf x| = 1}$, the graph of any unbounded function $f(x,y)$ defined over $Bbb R^2$. All of these are closed but not bounded! (as subsets of $Bbb R^3$)
– GPerez
Mar 15 '15 at 12:24
@user2715119 So it is "closed", in the sense of being closed under addition and multiplication (as well as satisfying the other axioms of vector spaces)? Not closed as in "interval containing its endpoints"? Since $mathbb{R}$ is not a closed interval, I assume you are referring to the first part? So a closed geometric shape is one where the set of all points contained "inside" the shape, is closed?
– Laplacinator
Mar 15 '15 at 12:40