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Domain to define a holomorphic function with non-isolated singularities closed, open, neither?
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I'm studying singularities in complex analysis and came across the following function:
$$f(z):=frac1zcos(frac1z)$$
I already figured out that $f$ has a non-isolated singularity in $0$ and that there are poles in $$z_k:=frac2pi(1+2k)$$ for $kinmathbbZ$, tending towards $0$ for $|k|toinfty$. I am now trying to find out if the domain $$D_f:=mathbbCsetminus(z_kcup0),$$ is closed, open or neither.
I already came up with a proof that $D_f$ cannot be closed because if $zinz_kcup0$, there is no $epsilon>0$ such that $B(z_1,epsilon)subseteq z_kcup0$ due to $z_kle z_1$ for all $kinmathbbZ$.
Can one help me out with the rest?
Thank you in advance.
general-topology complex-analysis
asked 2 days ago
physicist23
33
add a comment |Â
up vote
0
down vote
favorite
I'm studying singularities in complex analysis and came across the following function:
$$f(z):=frac1zcos(frac1z)$$
I already figured out that $f$ has a non-isolated singularity in $0$ and that there are poles in $$z_k:=frac2pi(1+2k)$$ for $kinmathbbZ$, tending towards $0$ for $|k|toinfty$. I am now trying to find out if the domain $$D_f:=mathbbCsetminus(z_kcup0),$$ is closed, open or neither.
I already came up with a proof that $D_f$ cannot be closed because if $zinz_kcup0$, there is no $epsilon>0$ such that $B(z_1,epsilon)subseteq z_kcup0$ due to $z_kle z_1$ for all $kinmathbbZ$.
Can one help me out with the rest?
Thank you in advance.
general-topology complex-analysis
asked 2 days ago
physicist23
33
First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm studying singularities in complex analysis and came across the following function:
$$f(z):=frac1zcos(frac1z)$$
I already figured out that $f$ has a non-isolated singularity in $0$ and that there are poles in $$z_k:=frac2pi(1+2k)$$ for $kinmathbbZ$, tending towards $0$ for $|k|toinfty$. I am now trying to find out if the domain $$D_f:=mathbbCsetminus(z_kcup0),$$ is closed, open or neither.
I already came up with a proof that $D_f$ cannot be closed because if $zinz_kcup0$, there is no $epsilon>0$ such that $B(z_1,epsilon)subseteq z_kcup0$ due to $z_kle z_1$ for all $kinmathbbZ$.
Can one help me out with the rest?
Thank you in advance.
general-topology complex-analysis
asked 2 days ago
physicist23
33
I'm studying singularities in complex analysis and came across the following function:
$$f(z):=frac1zcos(frac1z)$$
I already figured out that $f$ has a non-isolated singularity in $0$ and that there are poles in $$z_k:=frac2pi(1+2k)$$ for $kinmathbbZ$, tending towards $0$ for $|k|toinfty$. I am now trying to find out if the domain $$D_f:=mathbbCsetminus(z_kcup0),$$ is closed, open or neither.
I already came up with a proof that $D_f$ cannot be closed because if $zinz_kcup0$, there is no $epsilon>0$ such that $B(z_1,epsilon)subseteq z_kcup0$ due to $z_kle z_1$ for all $kinmathbbZ$.
Can one help me out with the rest?
Thank you in advance.
general-topology complex-analysis
asked 2 days ago
physicist23
33
edited 2 days ago
asked 2 days ago
physicist23
33
asked 2 days ago
physicist23
33
asked 2 days ago
physicist23
33
33
First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago
add a comment |Â
First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago
First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago
add a comment |Â
1 Answer
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$D_f = Bbb Cbackslash left( left z_k right_ - infty ^infty cup left 0 right right)$ is in fact open.
I believe it is easier to see that $B = left z_k right_ - infty ^infty cup left 0 right$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $$D_f = Bbb Cbackslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point.
Thus $B$ is closed.
answered 2 days ago
zokomoko
305213
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$D_f = Bbb Cbackslash left( left z_k right_ - infty ^infty cup left 0 right right)$ is in fact open.
I believe it is easier to see that $B = left z_k right_ - infty ^infty cup left 0 right$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $$D_f = Bbb Cbackslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point.
Thus $B$ is closed.
answered 2 days ago
zokomoko
305213
add a comment |Â
up vote
0
down vote
accepted
$D_f = Bbb Cbackslash left( left z_k right_ - infty ^infty cup left 0 right right)$ is in fact open.
I believe it is easier to see that $B = left z_k right_ - infty ^infty cup left 0 right$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $$D_f = Bbb Cbackslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point.
Thus $B$ is closed.
answered 2 days ago
zokomoko
305213
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$D_f = Bbb Cbackslash left( left z_k right_ - infty ^infty cup left 0 right right)$ is in fact open.
I believe it is easier to see that $B = left z_k right_ - infty ^infty cup left 0 right$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $$D_f = Bbb Cbackslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point.
Thus $B$ is closed.
answered 2 days ago
zokomoko
305213
$D_f = Bbb Cbackslash left( left z_k right_ - infty ^infty cup left 0 right right)$ is in fact open.
I believe it is easier to see that $B = left z_k right_ - infty ^infty cup left 0 right$ is closed.
According to this:
Prove that a set is closed iff it contains all its accumulation points
it suffices to check that $B$ contains all of its accumulation points.
Well the set $B$ has only one accumulation point and it's $0$ and it's contained in $B$.
Thus $B$ is closed and $$D_f = Bbb Cbackslash B$$ is open.
Alternatively we can show that the set $B$ is closed by showing that it contains all of its boundary points.
By the definition of a boundary point ($x$ is a boundary points of $B$ if every neighborhood of $x$ contains at least one element in $B$ and at least one element not in $B$) we conclude that every point in $B$ is a boundary point.
Thus $B$ is closed.
answered 2 days ago
zokomoko
305213
edited 2 days ago
answered 2 days ago
zokomoko
305213
answered 2 days ago
zokomoko
305213
answered 2 days ago
zokomoko
305213
305213
add a comment |Â
add a comment |Â
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First, 0 is not a pole but rather a non-isolated singularity...z_k are on the other hand simple poles. Second, why do you think D_f is not open?
– zokomoko
2 days ago
Thanks, I edited the question. Actually, I am lacking a proof here... Can one proof that it is open?
– physicist23
2 days ago