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)?










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  • 1




    Right, that (informally described) set is not open, hence not a domain (it is connected, however).
    – Daniel Fischer
    Aug 24 '13 at 20:17










  • However, it is closed correct?
    – Mr.Fry
    Aug 24 '13 at 20:42










  • Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
    – Daniel Fischer
    Aug 24 '13 at 20:44
















0














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)?










share|cite|improve this question


















  • 1




    Right, that (informally described) set is not open, hence not a domain (it is connected, however).
    – Daniel Fischer
    Aug 24 '13 at 20:17










  • However, it is closed correct?
    – Mr.Fry
    Aug 24 '13 at 20:42










  • Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
    – Daniel Fischer
    Aug 24 '13 at 20:44














0












0








0


1





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)?










share|cite|improve this question













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






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asked Aug 24 '13 at 20:14









Mr.FryMr.Fry

3,88021223




3,88021223








  • 1




    Right, that (informally described) set is not open, hence not a domain (it is connected, however).
    – Daniel Fischer
    Aug 24 '13 at 20:17










  • However, it is closed correct?
    – Mr.Fry
    Aug 24 '13 at 20:42










  • Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
    – Daniel Fischer
    Aug 24 '13 at 20:44














  • 1




    Right, that (informally described) set is not open, hence not a domain (it is connected, however).
    – Daniel Fischer
    Aug 24 '13 at 20:17










  • However, it is closed correct?
    – Mr.Fry
    Aug 24 '13 at 20:42










  • Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
    – Daniel Fischer
    Aug 24 '13 at 20:44








1




1




Right, that (informally described) set is not open, hence not a domain (it is connected, however).
– Daniel Fischer
Aug 24 '13 at 20:17




Right, that (informally described) set is not open, hence not a domain (it is connected, however).
– Daniel Fischer
Aug 24 '13 at 20:17












However, it is closed correct?
– Mr.Fry
Aug 24 '13 at 20:42




However, it is closed correct?
– Mr.Fry
Aug 24 '13 at 20:42












Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
– Daniel Fischer
Aug 24 '13 at 20:44




Yes, this one is closed. Just to make sure (you probably know it already well), most sets are neither open nor closed.
– Daniel Fischer
Aug 24 '13 at 20:44










3 Answers
3






active

oldest

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1














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.






share|cite|improve this answer





















  • 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?
    – Mr.Fry
    Aug 24 '13 at 20:33










  • @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
    – njguliyev
    Aug 24 '13 at 21:15










  • cool, I understand that.
    – Mr.Fry
    Aug 24 '13 at 21:21



















0














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.






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    0














    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.






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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      1














      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.






      share|cite|improve this answer





















      • 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?
        – Mr.Fry
        Aug 24 '13 at 20:33










      • @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
        – njguliyev
        Aug 24 '13 at 21:15










      • cool, I understand that.
        – Mr.Fry
        Aug 24 '13 at 21:21
















      1














      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.






      share|cite|improve this answer





















      • 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?
        – Mr.Fry
        Aug 24 '13 at 20:33










      • @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
        – njguliyev
        Aug 24 '13 at 21:15










      • cool, I understand that.
        – Mr.Fry
        Aug 24 '13 at 21:21














      1












      1








      1






      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.






      share|cite|improve this answer












      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.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Aug 24 '13 at 20:23







      user61527



















      • 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?
        – Mr.Fry
        Aug 24 '13 at 20:33










      • @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
        – njguliyev
        Aug 24 '13 at 21:15










      • cool, I understand that.
        – Mr.Fry
        Aug 24 '13 at 21:21


















      • 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?
        – Mr.Fry
        Aug 24 '13 at 20:33










      • @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
        – njguliyev
        Aug 24 '13 at 21:15










      • cool, I understand that.
        – Mr.Fry
        Aug 24 '13 at 21:21
















      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?
      – Mr.Fry
      Aug 24 '13 at 20:33




      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?
      – Mr.Fry
      Aug 24 '13 at 20:33












      @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
      – njguliyev
      Aug 24 '13 at 21:15




      @Crypto, no, this set is also connected, but again not open. In your division $x=1$ is not open.
      – njguliyev
      Aug 24 '13 at 21:15












      cool, I understand that.
      – Mr.Fry
      Aug 24 '13 at 21:21




      cool, I understand that.
      – Mr.Fry
      Aug 24 '13 at 21:21











      0














      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.






      share|cite|improve this answer


























        0














        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.






        share|cite|improve this answer
























          0












          0








          0






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 24 '13 at 20:24









          Owen SizemoreOwen Sizemore

          4,8641319




          4,8641319























              0














              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.






              share|cite|improve this answer


























                0














                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.






                share|cite|improve this answer
























                  0












                  0








                  0






                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 2:51









                  Duchamp Gérard H. E.Duchamp Gérard H. E.

                  2,552918




                  2,552918






























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