Complex Analysis - Let $Omega subseteq BbbC$ be a domain.
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I'm trying to prove the following.
"Let $Omega subseteq BbbC$ be a domain. and let $epsilon>0$. For any two points $u,zinOmega$, prove that there exists a sequence of points $w_i^N_i=1subseteqOmega$ satisfying $w_1=u$, $w_N=z$, and $|w_i -w_i+1|<epsilon$ for all $i=1,...,N-1$."
It is obvious to show that domain is path-connected by setting two disjoint open subsets and concatenate the path within the points to show that one of the open subsets is, in fact, the whole domain. However, I'm not fully understanding how to go by solving the actual proof where we are showing the two consecutive points have distance less than some $epsilon >0$
edit - Sorry I forgot to mention... I'm not supposed to use the path-connectedness
general-topology complex-analysis connectedness
asked Sep 1 at 1:08
Ya G
1629
add a comment |Â
up vote
1
down vote
favorite
I'm trying to prove the following.
"Let $Omega subseteq BbbC$ be a domain. and let $epsilon>0$. For any two points $u,zinOmega$, prove that there exists a sequence of points $w_i^N_i=1subseteqOmega$ satisfying $w_1=u$, $w_N=z$, and $|w_i -w_i+1|<epsilon$ for all $i=1,...,N-1$."
It is obvious to show that domain is path-connected by setting two disjoint open subsets and concatenate the path within the points to show that one of the open subsets is, in fact, the whole domain. However, I'm not fully understanding how to go by solving the actual proof where we are showing the two consecutive points have distance less than some $epsilon >0$
edit - Sorry I forgot to mention... I'm not supposed to use the path-connectedness
general-topology complex-analysis connectedness
asked Sep 1 at 1:08
Ya G
1629
1
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
1
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to prove the following.
"Let $Omega subseteq BbbC$ be a domain. and let $epsilon>0$. For any two points $u,zinOmega$, prove that there exists a sequence of points $w_i^N_i=1subseteqOmega$ satisfying $w_1=u$, $w_N=z$, and $|w_i -w_i+1|<epsilon$ for all $i=1,...,N-1$."
It is obvious to show that domain is path-connected by setting two disjoint open subsets and concatenate the path within the points to show that one of the open subsets is, in fact, the whole domain. However, I'm not fully understanding how to go by solving the actual proof where we are showing the two consecutive points have distance less than some $epsilon >0$
edit - Sorry I forgot to mention... I'm not supposed to use the path-connectedness
general-topology complex-analysis connectedness
asked Sep 1 at 1:08
Ya G
1629
I'm trying to prove the following.
"Let $Omega subseteq BbbC$ be a domain. and let $epsilon>0$. For any two points $u,zinOmega$, prove that there exists a sequence of points $w_i^N_i=1subseteqOmega$ satisfying $w_1=u$, $w_N=z$, and $|w_i -w_i+1|<epsilon$ for all $i=1,...,N-1$."
It is obvious to show that domain is path-connected by setting two disjoint open subsets and concatenate the path within the points to show that one of the open subsets is, in fact, the whole domain. However, I'm not fully understanding how to go by solving the actual proof where we are showing the two consecutive points have distance less than some $epsilon >0$
edit - Sorry I forgot to mention... I'm not supposed to use the path-connectedness
general-topology complex-analysis connectedness
general-topology complex-analysis connectedness
asked Sep 1 at 1:08
Ya G
1629
asked Sep 1 at 1:08
Ya G
1629
edited Sep 1 at 1:22
asked Sep 1 at 1:08
Ya G
1629
asked Sep 1 at 1:08
Ya G
1629
asked Sep 1 at 1:08
Ya G
1629
1629
1
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
1
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49
add a comment |Â
1
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
1
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49
1
1
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
1
1
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49
add a comment |Â
1 Answer
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up vote
1
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accepted
A typical argument to show that all elements of a connected set $Omega$ satisfy
a certain property $P$ is to show that both the subset of all points
satisfying $P$ and its complement are open in $Omega$. This works here
as well:
For fixed $u in Omega$ and $epsilon > 0$ define two sets
$$
A = w_i -w_i+1
$$
and its complement $B = Omega setminus A$.
Then show that both $A$ and $B$ are open sets: If
$z in A$ (resp. $B$) and $U_delta(z) subset Omega$ for some
$0 < delta le epsilon$, then $U_delta(z) subset A$ (resp. $B$).
($U_delta(z)$ denotes the open disc with center $z$ and radius
$delta$.)
Since $Omega$ is connected, one of them must be empty, and consequently,
$Omega = A$.
answered Sep 1 at 10:29
Martin R
24k32744
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A typical argument to show that all elements of a connected set $Omega$ satisfy
a certain property $P$ is to show that both the subset of all points
satisfying $P$ and its complement are open in $Omega$. This works here
as well:
For fixed $u in Omega$ and $epsilon > 0$ define two sets
$$
A = w_i -w_i+1
$$
and its complement $B = Omega setminus A$.
Then show that both $A$ and $B$ are open sets: If
$z in A$ (resp. $B$) and $U_delta(z) subset Omega$ for some
$0 < delta le epsilon$, then $U_delta(z) subset A$ (resp. $B$).
($U_delta(z)$ denotes the open disc with center $z$ and radius
$delta$.)
Since $Omega$ is connected, one of them must be empty, and consequently,
$Omega = A$.
answered Sep 1 at 10:29
Martin R
24k32744
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
 |Â
show 4 more comments
up vote
1
down vote
accepted
A typical argument to show that all elements of a connected set $Omega$ satisfy
a certain property $P$ is to show that both the subset of all points
satisfying $P$ and its complement are open in $Omega$. This works here
as well:
For fixed $u in Omega$ and $epsilon > 0$ define two sets
$$
A = w_i -w_i+1
$$
and its complement $B = Omega setminus A$.
Then show that both $A$ and $B$ are open sets: If
$z in A$ (resp. $B$) and $U_delta(z) subset Omega$ for some
$0 < delta le epsilon$, then $U_delta(z) subset A$ (resp. $B$).
($U_delta(z)$ denotes the open disc with center $z$ and radius
$delta$.)
Since $Omega$ is connected, one of them must be empty, and consequently,
$Omega = A$.
answered Sep 1 at 10:29
Martin R
24k32744
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A typical argument to show that all elements of a connected set $Omega$ satisfy
a certain property $P$ is to show that both the subset of all points
satisfying $P$ and its complement are open in $Omega$. This works here
as well:
For fixed $u in Omega$ and $epsilon > 0$ define two sets
$$
A = w_i -w_i+1
$$
and its complement $B = Omega setminus A$.
Then show that both $A$ and $B$ are open sets: If
$z in A$ (resp. $B$) and $U_delta(z) subset Omega$ for some
$0 < delta le epsilon$, then $U_delta(z) subset A$ (resp. $B$).
($U_delta(z)$ denotes the open disc with center $z$ and radius
$delta$.)
Since $Omega$ is connected, one of them must be empty, and consequently,
$Omega = A$.
answered Sep 1 at 10:29
Martin R
24k32744
A typical argument to show that all elements of a connected set $Omega$ satisfy
a certain property $P$ is to show that both the subset of all points
satisfying $P$ and its complement are open in $Omega$. This works here
as well:
For fixed $u in Omega$ and $epsilon > 0$ define two sets
$$
A = w_i -w_i+1
$$
and its complement $B = Omega setminus A$.
Then show that both $A$ and $B$ are open sets: If
$z in A$ (resp. $B$) and $U_delta(z) subset Omega$ for some
$0 < delta le epsilon$, then $U_delta(z) subset A$ (resp. $B$).
($U_delta(z)$ denotes the open disc with center $z$ and radius
$delta$.)
Since $Omega$ is connected, one of them must be empty, and consequently,
$Omega = A$.
answered Sep 1 at 10:29
Martin R
24k32744
edited Sep 1 at 10:36
answered Sep 1 at 10:29
Martin R
24k32744
answered Sep 1 at 10:29
Martin R
24k32744
answered Sep 1 at 10:29
Martin R
24k32744
24k32744
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
 |Â
show 4 more comments
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Here is my attempt. Let me know if I'm missing something. Since $Omega$ is open and connected, fix a point $uinOmega$ and $epsilon>0$, define $Omega_1:=<epsilontext for 1leq ileq N-1$ and $Omega_2=Omega-backslashOmega_1$. Evidently, $Omega_1cupOmega_2=Omega$ and $Omega_1capOmega_2=phi$. Let $w_iinOmega_1$ for some $iin[1,N-1].$
– Ya G
Sep 1 at 16:16
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Since $Omega$ is open, there exists $D_epsilon(w_i)subseteqOmega$ for some $epsilon>0$. It is obvious that if $w_i+1in D_epsilon(w_i)$, then $|w_i-w_i+1|<epsilon$. Then, consider the mapping defined by $pi(t)=begincases pi(2t) & 0leq t<frac12\ pi(2t-1) & frac12<tleq1 endcases$. Then, $pi$ is continuous, piecewise smooth curve that connects $w_i$ to $w_i+1$. It follows that $D_epsilon(w_i)subseteqOmega_1$ and that $Omega_1$ is open.
– Ya G
Sep 1 at 16:20
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
Now suppose $w_i+1inOmega_2$. Since $Omega$ is open, there exists $D_epsilon(w_i+1)subseteqOmega$. Let $Omega_i+2inOmega_2$. If there exists a curve that connected $w_i+1$ to $w_i+2$, then by taking similar approach as above, find a curve connecting $w_i$ to $w_i+2$ by concatenating the path from $w_i$ to $w_i+1$ and the path from $w_i+1$ to $w_i+2$. Thus, $w_i+2inOmega$, which is a contradiction to our initial assumption.
– Ya G
Sep 1 at 16:22
1
1
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
@YaG: Your $Omega_1, Omega_2$ are the $A, B$ from my answer. But there are no paths involved. If $z in A$ then there is a chain $u = w_1,w_2, ldots w_N-1, w_N = z$ connecting $u$ with $z$. If $D_delta(z) subset Omega$ then for each $z' in D_delta(z)$, $u = w_1,w_2, ldots w_N-1, w_N = z, w_N+1 = z'$ is a chain connecting $u$ with $z'$. Hence $D_delta(z) subset A$. Similarly for $B$. – Then the connectedness is used as you (and I) said.
– Martin R
Sep 1 at 16:28
1
1
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
@YaG: Yes, that's what I meant. One thing only: $epsilon > 0$ is given, so you cannot say that $D_epsilon(z)subseteqOmega$, only that $D_delta(z)subseteqOmega$ for some $0 < delta le epsilon$.
– Martin R
Sep 1 at 16:59
 |Â
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1
Hint: the path is compact in $mathbb C$.
– xbh
Sep 1 at 1:21
1
Use that the domain is connected. You problem is a special case of the topological chain theorem for clnnecged spaces.
– William Elliot
Sep 1 at 7:49