Understanding the definition of domain in Complex Analysis
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I have a definition in my book which states, "a nonempty open set that is connected is called a domain." I understand what an open set is (a set containing none of its boundary points and I know what a boundary point is). I am a bit confused with the definition of connected.
Ex. |z-3+2i| $ge$ 1. We can translate this to (x-3)^2+(y+2)^2 $ge$ 1. Is this not a domain because this set contains the boundary points of the circle centered at (3,-2)?
complex-analysis definition
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
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add a comment |
$begingroup$
I have a definition in my book which states, "a nonempty open set that is connected is called a domain." I understand what an open set is (a set containing none of its boundary points and I know what a boundary point is). I am a bit confused with the definition of connected.
Ex. |z-3+2i| $ge$ 1. We can translate this to (x-3)^2+(y+2)^2 $ge$ 1. Is this not a domain because this set contains the boundary points of the circle centered at (3,-2)?
complex-analysis definition
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
$endgroup$
1
$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
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However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
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Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44
add a comment |
$begingroup$
I have a definition in my book which states, "a nonempty open set that is connected is called a domain." I understand what an open set is (a set containing none of its boundary points and I know what a boundary point is). I am a bit confused with the definition of connected.
Ex. |z-3+2i| $ge$ 1. We can translate this to (x-3)^2+(y+2)^2 $ge$ 1. Is this not a domain because this set contains the boundary points of the circle centered at (3,-2)?
complex-analysis definition
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
$endgroup$
I have a definition in my book which states, "a nonempty open set that is connected is called a domain." I understand what an open set is (a set containing none of its boundary points and I know what a boundary point is). I am a bit confused with the definition of connected.
Ex. |z-3+2i| $ge$ 1. We can translate this to (x-3)^2+(y+2)^2 $ge$ 1. Is this not a domain because this set contains the boundary points of the circle centered at (3,-2)?
complex-analysis definition
complex-analysis definition
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
asked Aug 24 '13 at 20:14
Mr.FryMr.Fry
3,89021323
3,89021323
1
$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
$begingroup$
However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
$begingroup$
Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44
add a comment |
1
$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
$begingroup$
However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
$begingroup$
Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44
1
1
$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
$begingroup$
However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
$begingroup$
However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
$begingroup$
Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44
$begingroup$
Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
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Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open.
$endgroup$
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
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@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
add a comment |
$begingroup$
In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
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add a comment |
$begingroup$
I think the deep meaning of "Domain $Dsubset mathbb{C}$" is that you cannot write $D=D_1sqcup D_2, D_inot=emptyset$ (disjoint union). Indeed, if $Dsubset mathbb{C}$ is only open, consider the algebra $mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1sqcup D_2, D_inot=emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}not=0$. Hope it helps.
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open.
$endgroup$
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
add a comment |
$begingroup$
Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open.
$endgroup$
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
add a comment |
$begingroup$
Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open.
$endgroup$
Formally, connected means that we cannot break the domain up into two disjoint non-empty open sets. The picture you should have in mind is a region that is "all one piece."
So the example you gave is connected (it's the entire plane except for an open disk of radius $1$ around the point $3 - 2i$) but it's not a domain, since it's not open.
answered Aug 24 '13 at 20:23
user61527
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
add a comment |
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
So for another example |z-2|$ge$ |z|, which can be translated to x$leq$ 1 is not connected because the two disjoint subsets of the domain are x=1, x<1?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:33
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
@Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
$endgroup$
– njguliyev
Aug 24 '13 at 21:15
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
$begingroup$
cool, I understand that.
$endgroup$
– Mr.Fry
Aug 24 '13 at 21:21
add a comment |
$begingroup$
In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
$endgroup$
add a comment |
$begingroup$
In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
$endgroup$
add a comment |
$begingroup$
In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
$endgroup$
In this situation a domain is also path-connected (this means that given any two points in the domain you can connect them by a path that stays in the domain). So the intuitive picture that you should is to draw a curve that a.) does not intersect itself b.) cuts the plane into 2 pieces.
If you think about it there are two ways to do this.
1.) Draw a closed loop.
2.) Draw a curve such that both ends go off to infinity at some point
(Note: there are some other issues like drawing a curve that looks like a number 6 but that actually work for what I am saying, and to get rid of this case you can only consider curves that extend to embeddings of $S^1$ in the Riemann sphere)
Then if we look at the two parts of the plane and DON'T include the curve these will be domains. This is the type of picture you should have in your head. Though not all domains look exactly like this, they are pretty close.
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
answered Aug 24 '13 at 20:24
Owen SizemoreOwen Sizemore
4,9641319
4,9641319
add a comment |
add a comment |
$begingroup$
I think the deep meaning of "Domain $Dsubset mathbb{C}$" is that you cannot write $D=D_1sqcup D_2, D_inot=emptyset$ (disjoint union). Indeed, if $Dsubset mathbb{C}$ is only open, consider the algebra $mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1sqcup D_2, D_inot=emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}not=0$. Hope it helps.
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
I think the deep meaning of "Domain $Dsubset mathbb{C}$" is that you cannot write $D=D_1sqcup D_2, D_inot=emptyset$ (disjoint union). Indeed, if $Dsubset mathbb{C}$ is only open, consider the algebra $mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1sqcup D_2, D_inot=emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}not=0$. Hope it helps.
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
add a comment |
$begingroup$
I think the deep meaning of "Domain $Dsubset mathbb{C}$" is that you cannot write $D=D_1sqcup D_2, D_inot=emptyset$ (disjoint union). Indeed, if $Dsubset mathbb{C}$ is only open, consider the algebra $mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1sqcup D_2, D_inot=emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}not=0$. Hope it helps.
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
$endgroup$
I think the deep meaning of "Domain $Dsubset mathbb{C}$" is that you cannot write $D=D_1sqcup D_2, D_inot=emptyset$ (disjoint union). Indeed, if $Dsubset mathbb{C}$ is only open, consider the algebra $mathcal{H}(D)$ of analytic functions on it. This algebra is a domain (in the algebraic sense, i.e. without zero divisor) iff $D$ is connected as if $D=D_1sqcup D_2, D_inot=emptyset$, we have $0_D=1_{D_1}.1_{D_2}$ and $1_{D_i}not=0$. Hope it helps.
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
answered Jan 4 at 2:51
Duchamp Gérard H. E.Duchamp Gérard H. E.
2,644919
2,644919
add a comment |
add a comment |
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$begingroup$
Right, that (informally described) set is not open, hence not a domain (it is connected, however).
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:17
$begingroup$
However, it is closed correct?
$endgroup$
– Mr.Fry
Aug 24 '13 at 20:42
$begingroup$
Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
$endgroup$
– Daniel Fischer
Aug 24 '13 at 20:44