Is the domain of a complex function always open? Is $mathbb C$ (the domain) always of measure zero? What if...












0












$begingroup$


Usually the domain of a complex function is $mathbb Cbackslash{zinmathbb C~:~z text{ is a singularity of } f }$ So I guess it must always be $mathbb Cbackslash{text{a set of points}}$. Moreover it's usually a finite or at most countable set of points.



So the domain doesn't need to be open because it can be the infinite intersection $bigcaplimits_{iin A}{}mathbb Cbackslash{z_i}$ and $A$ counld be infinite. For instance $bigcaplimits_{ninBbb N}{}mathbb Cbackslash{{1over n}}$ is not open.



Can the domain be closed?



Is there a complex function that is not defined on a connected set that is more than a segment?



What if $f$ is holomorphic/meromorphic?






complex-analysis holomorphic-functions meromorphic-functions







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$endgroup$












  • $begingroup$
    Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 10:26
















0












$begingroup$


Usually the domain of a complex function is $mathbb Cbackslash{zinmathbb C~:~z text{ is a singularity of } f }$ So I guess it must always be $mathbb Cbackslash{text{a set of points}}$. Moreover it's usually a finite or at most countable set of points.



So the domain doesn't need to be open because it can be the infinite intersection $bigcaplimits_{iin A}{}mathbb Cbackslash{z_i}$ and $A$ counld be infinite. For instance $bigcaplimits_{ninBbb N}{}mathbb Cbackslash{{1over n}}$ is not open.



Can the domain be closed?



Is there a complex function that is not defined on a connected set that is more than a segment?



What if $f$ is holomorphic/meromorphic?






complex-analysis holomorphic-functions meromorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 10:26














0












0








0





$begingroup$


Usually the domain of a complex function is $mathbb Cbackslash{zinmathbb C~:~z text{ is a singularity of } f }$ So I guess it must always be $mathbb Cbackslash{text{a set of points}}$. Moreover it's usually a finite or at most countable set of points.



So the domain doesn't need to be open because it can be the infinite intersection $bigcaplimits_{iin A}{}mathbb Cbackslash{z_i}$ and $A$ counld be infinite. For instance $bigcaplimits_{ninBbb N}{}mathbb Cbackslash{{1over n}}$ is not open.



Can the domain be closed?



Is there a complex function that is not defined on a connected set that is more than a segment?



What if $f$ is holomorphic/meromorphic?






complex-analysis holomorphic-functions meromorphic-functions







share|cite|improve this question











$endgroup$




Usually the domain of a complex function is $mathbb Cbackslash{zinmathbb C~:~z text{ is a singularity of } f }$ So I guess it must always be $mathbb Cbackslash{text{a set of points}}$. Moreover it's usually a finite or at most countable set of points.



So the domain doesn't need to be open because it can be the infinite intersection $bigcaplimits_{iin A}{}mathbb Cbackslash{z_i}$ and $A$ counld be infinite. For instance $bigcaplimits_{ninBbb N}{}mathbb Cbackslash{{1over n}}$ is not open.



Can the domain be closed?



Is there a complex function that is not defined on a connected set that is more than a segment?



What if $f$ is holomorphic/meromorphic?






complex-analysis holomorphic-functions meromorphic-functions




complex-analysis holomorphic-functions meromorphic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 11:27







John Cataldo

















asked Dec 31 '18 at 10:16









John CataldoJohn Cataldo

1,1881316




1,1881316












  • $begingroup$
    Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 10:26


















  • $begingroup$
    Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
    $endgroup$
    – Kavi Rama Murthy
    Dec 31 '18 at 10:26
















$begingroup$
Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:26




$begingroup$
Functions can be defined from any set to any other set. For holomorphic/ meromorphic functions the domain is taken to be open.
$endgroup$
– Kavi Rama Murthy
Dec 31 '18 at 10:26










1 Answer
1






active

oldest

votes


















0












$begingroup$

First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.



It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.



On the other hand, if $rin(0,1]$ and if I define$$begin{array}{rccc}fcolon&D(0,r)&longrightarrow&mathbb C\&z&mapsto&displaystylesum_{n=0}^infty z^n,end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:22












  • $begingroup$
    Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:40










  • $begingroup$
    Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:42










  • $begingroup$
    I suppose that what you have in mind is analytic continuation.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:47










  • $begingroup$
    Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:49











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.



It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.



On the other hand, if $rin(0,1]$ and if I define$$begin{array}{rccc}fcolon&D(0,r)&longrightarrow&mathbb C\&z&mapsto&displaystylesum_{n=0}^infty z^n,end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:22












  • $begingroup$
    Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:40










  • $begingroup$
    Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:42










  • $begingroup$
    I suppose that what you have in mind is analytic continuation.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:47










  • $begingroup$
    Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:49
















0












$begingroup$

First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.



It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.



On the other hand, if $rin(0,1]$ and if I define$$begin{array}{rccc}fcolon&D(0,r)&longrightarrow&mathbb C\&z&mapsto&displaystylesum_{n=0}^infty z^n,end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:22












  • $begingroup$
    Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:40










  • $begingroup$
    Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:42










  • $begingroup$
    I suppose that what you have in mind is analytic continuation.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:47










  • $begingroup$
    Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:49














0












0








0





$begingroup$

First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.



It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.



On the other hand, if $rin(0,1]$ and if I define$$begin{array}{rccc}fcolon&D(0,r)&longrightarrow&mathbb C\&z&mapsto&displaystylesum_{n=0}^infty z^n,end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.






share|cite|improve this answer









$endgroup$



First of of all, asserting that the domain of a function $f$ is the complement of the set of singularities of $f$ makes no sense, because it is clearly a circular defintion.



It is just a matter of definition. Often, but not always, , the domain of a holomorphic function is open by definition.



On the other hand, if $rin(0,1]$ and if I define$$begin{array}{rccc}fcolon&D(0,r)&longrightarrow&mathbb C\&z&mapsto&displaystylesum_{n=0}^infty z^n,end{array}$$then the domain of $f$ is $D(0,r)$. This has nothing to do with singularities. And, yes, the domain can also be $overline{D(0,r)}$ (if you are not assuming that the domain is open by definition), unless $r=1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 10:28









José Carlos SantosJosé Carlos Santos

164k22131234




164k22131234












  • $begingroup$
    Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:22












  • $begingroup$
    Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:40










  • $begingroup$
    Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:42










  • $begingroup$
    I suppose that what you have in mind is analytic continuation.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:47










  • $begingroup$
    Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:49


















  • $begingroup$
    Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:22












  • $begingroup$
    Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:40










  • $begingroup$
    Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:42










  • $begingroup$
    I suppose that what you have in mind is analytic continuation.
    $endgroup$
    – José Carlos Santos
    Dec 31 '18 at 11:47










  • $begingroup$
    Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
    $endgroup$
    – John Cataldo
    Dec 31 '18 at 11:49
















$begingroup$
Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
$endgroup$
– John Cataldo
Dec 31 '18 at 11:22






$begingroup$
Isn't the domain defined as the set of points for which $f$ is defined? Is there a function that is not defined on a connected set that is more than a segment?
$endgroup$
– John Cataldo
Dec 31 '18 at 11:22














$begingroup$
Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
$endgroup$
– José Carlos Santos
Dec 31 '18 at 11:40




$begingroup$
Concerning your first question, the answer is negative. You must declare which is your domain (and codomain) first and only then to declare state what $f$ is.
$endgroup$
– José Carlos Santos
Dec 31 '18 at 11:40












$begingroup$
Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
$endgroup$
– John Cataldo
Dec 31 '18 at 11:42




$begingroup$
Is there a word for the domain of definition? I'm interested in $f$ seen in the biggest possible space
$endgroup$
– John Cataldo
Dec 31 '18 at 11:42












$begingroup$
I suppose that what you have in mind is analytic continuation.
$endgroup$
– José Carlos Santos
Dec 31 '18 at 11:47




$begingroup$
I suppose that what you have in mind is analytic continuation.
$endgroup$
– José Carlos Santos
Dec 31 '18 at 11:47












$begingroup$
Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
$endgroup$
– John Cataldo
Dec 31 '18 at 11:49




$begingroup$
Yes. Is is true that any function can be extended by analytical continuation to almost $mathbb C$?
$endgroup$
– John Cataldo
Dec 31 '18 at 11:49


















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