Mixing
Contniuty in $mathbb{C}$ of a function $f:Drightarrowmathbb{C},Dsubseteq mathbb{C}$
$begingroup$
$f:Drightarrowmathbb{C}$ and $Dsubseteq mathbb{C}$
It used to be a Question but now I think I have solved the Problem now.
I have sometimes confused the terms preimage and preimageset
Show that a function is continious in a point $z_0$ in $D$ if and only
if the preimages of neighbourhoods of $f(z_0)$ are neighbourhoods of
$z_0$ in $D$ and a function is continious if and only if the preimage
of every open subset in $ mathbb{C}$ is open in $ D$
Suppose $f$ is continious in $z_0$ if $|x-f(z_0)|<epsilon$ than it is necessary that $f^-1(x)$ is in a delta environment of $z_0$ otherwise it has no preimage.
Now the other side
LoveTooNap29 suggests to consider an arbitrary epsillon neighbourhood of $f(z_0)$, the preimage is again a neighbourhood of $z_0$ by assumption, neighbourhood of $z_0$ means that the Elements do not leave $D$ AND there exists a $rinmathbb{R}$ such that the distance between $z_0$ and any Point of the preimageset is smaller than $r$ which is equivalent to the existince of a deltaenvironment where all the preimages of the Elements in the neighbourhood of $f(z_0)$ sits in. Therefor picking a delta smaller or equal than $r$ and bigger then $0$ yields Images in lessthan-Epsilon distance of $f(z_0)$. Which is the Definition of continiuty written out.
But what is with the other Statement?
My Idea is to prove the first implication of the second Statement with a constructive Argument. f is continious and now I pick a open subset of $mathbb{C}$ that means two Things I cannot pick $mathbb{C}$ itself because then the corresponding immageset would be the whole set $f(D)$ which is never open. (I can also not pick a subset which contains $f(D)$). Also if I pick a subset where no Images of $f$ are within then the corresponding preimageset would also be empty and thus always open again by Definition. I am therefor picking a set which does not fall under those categories.
Because the set I have picked is open it means that it is a neighbourhood of every Point. That means it is a neighbourhood for every Point which has a preimage in $D$. Then applying what I have proved above and the fact that union of open sets are open again proves that the preimage of the open subset in $mathbb{C}$ we have chosen is also open in $D$.
Now assuming the other side : The preimage of every open subset in $mathbb{C}$ is also open $D$. What I Need to Show is that f is continious in every point.
Assuming there exists a point $z_0$ in $D$ where $f$ is not continious then there exists a neighbourhood of $f(z_0)$ which is a open subset of $mathbb{C}$ such that the preimageset $P$ is not a neighbourhood of $z_0$ (with preimageset I mean the set of all Elements in the neighbourhood of $f(z_0)$ which have a preimage). No neighbourhood means in this case that $forall_{rinmathbb{R}}exists_{p_r in P}|z_0-p|geq r$. Now $z_0in P$ is always true because a neighbourhood of $f(z_0)$ is a set which always contains $f(z_0)$. Since $P$ is not a neighbourhood of $z_0$, it cannot be open. This is a contradiction to the assumption.
$square$
complex-analysis
asked Jan 24 at 17:31
New2MathNew2Math
12312
$endgroup$
add a comment |
$begingroup$
$f:Drightarrowmathbb{C}$ and $Dsubseteq mathbb{C}$
It used to be a Question but now I think I have solved the Problem now.
I have sometimes confused the terms preimage and preimageset
Show that a function is continious in a point $z_0$ in $D$ if and only
if the preimages of neighbourhoods of $f(z_0)$ are neighbourhoods of
$z_0$ in $D$ and a function is continious if and only if the preimage
of every open subset in $ mathbb{C}$ is open in $ D$
Suppose $f$ is continious in $z_0$ if $|x-f(z_0)|<epsilon$ than it is necessary that $f^-1(x)$ is in a delta environment of $z_0$ otherwise it has no preimage.
Now the other side
LoveTooNap29 suggests to consider an arbitrary epsillon neighbourhood of $f(z_0)$, the preimage is again a neighbourhood of $z_0$ by assumption, neighbourhood of $z_0$ means that the Elements do not leave $D$ AND there exists a $rinmathbb{R}$ such that the distance between $z_0$ and any Point of the preimageset is smaller than $r$ which is equivalent to the existince of a deltaenvironment where all the preimages of the Elements in the neighbourhood of $f(z_0)$ sits in. Therefor picking a delta smaller or equal than $r$ and bigger then $0$ yields Images in lessthan-Epsilon distance of $f(z_0)$. Which is the Definition of continiuty written out.
But what is with the other Statement?
My Idea is to prove the first implication of the second Statement with a constructive Argument. f is continious and now I pick a open subset of $mathbb{C}$ that means two Things I cannot pick $mathbb{C}$ itself because then the corresponding immageset would be the whole set $f(D)$ which is never open. (I can also not pick a subset which contains $f(D)$). Also if I pick a subset where no Images of $f$ are within then the corresponding preimageset would also be empty and thus always open again by Definition. I am therefor picking a set which does not fall under those categories.
Because the set I have picked is open it means that it is a neighbourhood of every Point. That means it is a neighbourhood for every Point which has a preimage in $D$. Then applying what I have proved above and the fact that union of open sets are open again proves that the preimage of the open subset in $mathbb{C}$ we have chosen is also open in $D$.
Now assuming the other side : The preimage of every open subset in $mathbb{C}$ is also open $D$. What I Need to Show is that f is continious in every point.
Assuming there exists a point $z_0$ in $D$ where $f$ is not continious then there exists a neighbourhood of $f(z_0)$ which is a open subset of $mathbb{C}$ such that the preimageset $P$ is not a neighbourhood of $z_0$ (with preimageset I mean the set of all Elements in the neighbourhood of $f(z_0)$ which have a preimage). No neighbourhood means in this case that $forall_{rinmathbb{R}}exists_{p_r in P}|z_0-p|geq r$. Now $z_0in P$ is always true because a neighbourhood of $f(z_0)$ is a set which always contains $f(z_0)$. Since $P$ is not a neighbourhood of $z_0$, it cannot be open. This is a contradiction to the assumption.
$square$
complex-analysis
asked Jan 24 at 17:31
New2MathNew2Math
12312
$endgroup$
1
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
3
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50
add a comment |
$begingroup$
$f:Drightarrowmathbb{C}$ and $Dsubseteq mathbb{C}$
It used to be a Question but now I think I have solved the Problem now.
I have sometimes confused the terms preimage and preimageset
Show that a function is continious in a point $z_0$ in $D$ if and only
if the preimages of neighbourhoods of $f(z_0)$ are neighbourhoods of
$z_0$ in $D$ and a function is continious if and only if the preimage
of every open subset in $ mathbb{C}$ is open in $ D$
Suppose $f$ is continious in $z_0$ if $|x-f(z_0)|<epsilon$ than it is necessary that $f^-1(x)$ is in a delta environment of $z_0$ otherwise it has no preimage.
Now the other side
LoveTooNap29 suggests to consider an arbitrary epsillon neighbourhood of $f(z_0)$, the preimage is again a neighbourhood of $z_0$ by assumption, neighbourhood of $z_0$ means that the Elements do not leave $D$ AND there exists a $rinmathbb{R}$ such that the distance between $z_0$ and any Point of the preimageset is smaller than $r$ which is equivalent to the existince of a deltaenvironment where all the preimages of the Elements in the neighbourhood of $f(z_0)$ sits in. Therefor picking a delta smaller or equal than $r$ and bigger then $0$ yields Images in lessthan-Epsilon distance of $f(z_0)$. Which is the Definition of continiuty written out.
But what is with the other Statement?
My Idea is to prove the first implication of the second Statement with a constructive Argument. f is continious and now I pick a open subset of $mathbb{C}$ that means two Things I cannot pick $mathbb{C}$ itself because then the corresponding immageset would be the whole set $f(D)$ which is never open. (I can also not pick a subset which contains $f(D)$). Also if I pick a subset where no Images of $f$ are within then the corresponding preimageset would also be empty and thus always open again by Definition. I am therefor picking a set which does not fall under those categories.
Because the set I have picked is open it means that it is a neighbourhood of every Point. That means it is a neighbourhood for every Point which has a preimage in $D$. Then applying what I have proved above and the fact that union of open sets are open again proves that the preimage of the open subset in $mathbb{C}$ we have chosen is also open in $D$.
Now assuming the other side : The preimage of every open subset in $mathbb{C}$ is also open $D$. What I Need to Show is that f is continious in every point.
Assuming there exists a point $z_0$ in $D$ where $f$ is not continious then there exists a neighbourhood of $f(z_0)$ which is a open subset of $mathbb{C}$ such that the preimageset $P$ is not a neighbourhood of $z_0$ (with preimageset I mean the set of all Elements in the neighbourhood of $f(z_0)$ which have a preimage). No neighbourhood means in this case that $forall_{rinmathbb{R}}exists_{p_r in P}|z_0-p|geq r$. Now $z_0in P$ is always true because a neighbourhood of $f(z_0)$ is a set which always contains $f(z_0)$. Since $P$ is not a neighbourhood of $z_0$, it cannot be open. This is a contradiction to the assumption.
$square$
complex-analysis
asked Jan 24 at 17:31
New2MathNew2Math
12312
$endgroup$
$f:Drightarrowmathbb{C}$ and $Dsubseteq mathbb{C}$
It used to be a Question but now I think I have solved the Problem now.
I have sometimes confused the terms preimage and preimageset
Show that a function is continious in a point $z_0$ in $D$ if and only
if the preimages of neighbourhoods of $f(z_0)$ are neighbourhoods of
$z_0$ in $D$ and a function is continious if and only if the preimage
of every open subset in $ mathbb{C}$ is open in $ D$
Suppose $f$ is continious in $z_0$ if $|x-f(z_0)|<epsilon$ than it is necessary that $f^-1(x)$ is in a delta environment of $z_0$ otherwise it has no preimage.
Now the other side
LoveTooNap29 suggests to consider an arbitrary epsillon neighbourhood of $f(z_0)$, the preimage is again a neighbourhood of $z_0$ by assumption, neighbourhood of $z_0$ means that the Elements do not leave $D$ AND there exists a $rinmathbb{R}$ such that the distance between $z_0$ and any Point of the preimageset is smaller than $r$ which is equivalent to the existince of a deltaenvironment where all the preimages of the Elements in the neighbourhood of $f(z_0)$ sits in. Therefor picking a delta smaller or equal than $r$ and bigger then $0$ yields Images in lessthan-Epsilon distance of $f(z_0)$. Which is the Definition of continiuty written out.
But what is with the other Statement?
My Idea is to prove the first implication of the second Statement with a constructive Argument. f is continious and now I pick a open subset of $mathbb{C}$ that means two Things I cannot pick $mathbb{C}$ itself because then the corresponding immageset would be the whole set $f(D)$ which is never open. (I can also not pick a subset which contains $f(D)$). Also if I pick a subset where no Images of $f$ are within then the corresponding preimageset would also be empty and thus always open again by Definition. I am therefor picking a set which does not fall under those categories.
Because the set I have picked is open it means that it is a neighbourhood of every Point. That means it is a neighbourhood for every Point which has a preimage in $D$. Then applying what I have proved above and the fact that union of open sets are open again proves that the preimage of the open subset in $mathbb{C}$ we have chosen is also open in $D$.
Now assuming the other side : The preimage of every open subset in $mathbb{C}$ is also open $D$. What I Need to Show is that f is continious in every point.
Assuming there exists a point $z_0$ in $D$ where $f$ is not continious then there exists a neighbourhood of $f(z_0)$ which is a open subset of $mathbb{C}$ such that the preimageset $P$ is not a neighbourhood of $z_0$ (with preimageset I mean the set of all Elements in the neighbourhood of $f(z_0)$ which have a preimage). No neighbourhood means in this case that $forall_{rinmathbb{R}}exists_{p_r in P}|z_0-p|geq r$. Now $z_0in P$ is always true because a neighbourhood of $f(z_0)$ is a set which always contains $f(z_0)$. Since $P$ is not a neighbourhood of $z_0$, it cannot be open. This is a contradiction to the assumption.
$square$
complex-analysis
complex-analysis
asked Jan 24 at 17:31
New2MathNew2Math
12312
asked Jan 24 at 17:31
New2MathNew2Math
12312
edited Jan 24 at 19:39
New2Math
asked Jan 24 at 17:31
New2MathNew2Math
12312
asked Jan 24 at 17:31
New2MathNew2Math
12312
asked Jan 24 at 17:31
New2MathNew2Math
12312
12312
1
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
3
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50
add a comment |
1
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
3
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50
1
1
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
3
3
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50
add a comment |
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1
$begingroup$
What definition of continuity are you working with?
$endgroup$
– 0x539
Jan 24 at 17:33
$begingroup$
f is continious in z 0 then for all ϵ>0 there exists a δ>0 such that |z−z 0 |<ϵ⇒|f(z)−f(z 0 )|<ϵ
$endgroup$
– New2Math
Jan 24 at 17:35
$begingroup$
Pick an arbitrary neighborhood of $f(z_0)$. Consider it's pre-image. What do you notice?
$endgroup$
– LoveTooNap29
Jan 24 at 17:49
3
$begingroup$
Hint: This is true for arbitrary metric spaces. There's nothing special about $mathbb{C}$.
$endgroup$
– Connor Harris
Jan 24 at 17:50