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Elementary topology of $mathbb C$: Union of 2 regions with nonempty intersection is a region
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.31,32
Perhaps related but my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis: union-of-connected-subsets-is-connected-if-intersection-is-nonempty, Union of connected subsets is connected if intersection is nonempty
Pf:
Let $A$ and $B$ be the regions in question.
Their union is open is because they are each open. If that warrants further justification, then: any element in their union has a disc centred at the element that is contained entirely in A or in B and hence in A or B.
Suppose on the contrary that their union is not connected. Then $exists C_1, C_2, C_3, C_4$ s.t. $C_1 cup C_2 = A cup B, C_1 ne emptyset ne C_2, C_3 = C_3^0, C_4=C_4^0, C_1 subseteq C_3, C_2 subseteq C_4$ and $C_3 cap C_4 = emptyset$.
We deduce that $emptyset = C_1 cap C_2$, and so we have 4 cases:
Case 1: $A cup B subseteq C_1$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 2: $A cup B subseteq C_2$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 3: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
Case 4: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
The cases are exhaustive and each lead to a contradiction. $therefore,$ their union is connected. QED
For the former my proof is:
Definitions:
$$z in partial A iff exists r>0:D[z,r]cap A ne emptyset ne D[z,r]cap A^c$$
$$z in partial B iff exists r>0:D[z,r]cap B ne emptyset ne D[z,r]cap B^c$$
$$B textis closed iff partial B subseteq B iff B = B^0 cup partial B$$
Now observe for $D[z,r]cap A$:
$$D[z,r]cap A subseteq A subseteq B implies D[z,r]cap B ne emptyset$$
Thus for $D[z,r]cap A^c$:
$$D[z,r]cap A^c subseteq B iff D[z,r] subseteq B iff z in B^0$$
$$D[z,r]cap A^c subsetneq B iff D[z,r]cap B^c ne emptyset iff z in partial B$$
In either case $z in B$. QED
For the latter my proof is:
$$forall z in A=A^0, exists a>0: D[z,a] subset A =A^0 subseteq B$$
Now $$z in B^0 iff exists b>0:D[z,b] subseteq B$$
Choose $b=a$. QED
Where have I gone wrong?
real-analysis general-topology complex-analysis connectedness
asked Jul 29 at 11:34
BCLC
6,95921973
add a comment |Â
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.31,32
Perhaps related but my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis: union-of-connected-subsets-is-connected-if-intersection-is-nonempty, Union of connected subsets is connected if intersection is nonempty
Pf:
Let $A$ and $B$ be the regions in question.
Their union is open is because they are each open. If that warrants further justification, then: any element in their union has a disc centred at the element that is contained entirely in A or in B and hence in A or B.
Suppose on the contrary that their union is not connected. Then $exists C_1, C_2, C_3, C_4$ s.t. $C_1 cup C_2 = A cup B, C_1 ne emptyset ne C_2, C_3 = C_3^0, C_4=C_4^0, C_1 subseteq C_3, C_2 subseteq C_4$ and $C_3 cap C_4 = emptyset$.
We deduce that $emptyset = C_1 cap C_2$, and so we have 4 cases:
Case 1: $A cup B subseteq C_1$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 2: $A cup B subseteq C_2$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 3: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
Case 4: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
The cases are exhaustive and each lead to a contradiction. $therefore,$ their union is connected. QED
For the former my proof is:
Definitions:
$$z in partial A iff exists r>0:D[z,r]cap A ne emptyset ne D[z,r]cap A^c$$
$$z in partial B iff exists r>0:D[z,r]cap B ne emptyset ne D[z,r]cap B^c$$
$$B textis closed iff partial B subseteq B iff B = B^0 cup partial B$$
Now observe for $D[z,r]cap A$:
$$D[z,r]cap A subseteq A subseteq B implies D[z,r]cap B ne emptyset$$
Thus for $D[z,r]cap A^c$:
$$D[z,r]cap A^c subseteq B iff D[z,r] subseteq B iff z in B^0$$
$$D[z,r]cap A^c subsetneq B iff D[z,r]cap B^c ne emptyset iff z in partial B$$
In either case $z in B$. QED
For the latter my proof is:
$$forall z in A=A^0, exists a>0: D[z,a] subset A =A^0 subseteq B$$
Now $$z in B^0 iff exists b>0:D[z,b] subseteq B$$
Choose $b=a$. QED
Where have I gone wrong?
real-analysis general-topology complex-analysis connectedness
asked Jul 29 at 11:34
BCLC
6,95921973
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.31,32
Perhaps related but my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis: union-of-connected-subsets-is-connected-if-intersection-is-nonempty, Union of connected subsets is connected if intersection is nonempty
Pf:
Let $A$ and $B$ be the regions in question.
Their union is open is because they are each open. If that warrants further justification, then: any element in their union has a disc centred at the element that is contained entirely in A or in B and hence in A or B.
Suppose on the contrary that their union is not connected. Then $exists C_1, C_2, C_3, C_4$ s.t. $C_1 cup C_2 = A cup B, C_1 ne emptyset ne C_2, C_3 = C_3^0, C_4=C_4^0, C_1 subseteq C_3, C_2 subseteq C_4$ and $C_3 cap C_4 = emptyset$.
We deduce that $emptyset = C_1 cap C_2$, and so we have 4 cases:
Case 1: $A cup B subseteq C_1$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 2: $A cup B subseteq C_2$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 3: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
Case 4: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
The cases are exhaustive and each lead to a contradiction. $therefore,$ their union is connected. QED
For the former my proof is:
Definitions:
$$z in partial A iff exists r>0:D[z,r]cap A ne emptyset ne D[z,r]cap A^c$$
$$z in partial B iff exists r>0:D[z,r]cap B ne emptyset ne D[z,r]cap B^c$$
$$B textis closed iff partial B subseteq B iff B = B^0 cup partial B$$
Now observe for $D[z,r]cap A$:
$$D[z,r]cap A subseteq A subseteq B implies D[z,r]cap B ne emptyset$$
Thus for $D[z,r]cap A^c$:
$$D[z,r]cap A^c subseteq B iff D[z,r] subseteq B iff z in B^0$$
$$D[z,r]cap A^c subsetneq B iff D[z,r]cap B^c ne emptyset iff z in partial B$$
In either case $z in B$. QED
For the latter my proof is:
$$forall z in A=A^0, exists a>0: D[z,a] subset A =A^0 subseteq B$$
Now $$z in B^0 iff exists b>0:D[z,b] subseteq B$$
Choose $b=a$. QED
Where have I gone wrong?
real-analysis general-topology complex-analysis connectedness
asked Jul 29 at 11:34
BCLC
6,95921973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.31,32
Perhaps related but my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis: union-of-connected-subsets-is-connected-if-intersection-is-nonempty, Union of connected subsets is connected if intersection is nonempty
Pf:
Let $A$ and $B$ be the regions in question.
Their union is open is because they are each open. If that warrants further justification, then: any element in their union has a disc centred at the element that is contained entirely in A or in B and hence in A or B.
Suppose on the contrary that their union is not connected. Then $exists C_1, C_2, C_3, C_4$ s.t. $C_1 cup C_2 = A cup B, C_1 ne emptyset ne C_2, C_3 = C_3^0, C_4=C_4^0, C_1 subseteq C_3, C_2 subseteq C_4$ and $C_3 cap C_4 = emptyset$.
We deduce that $emptyset = C_1 cap C_2$, and so we have 4 cases:
Case 1: $A cup B subseteq C_1$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 2: $A cup B subseteq C_2$
$A cup B subset$ but $ne C_1 cup C_2$ ↯
Case 3: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
Case 4: $A subseteq C_1, B subseteq C_2$
$A cap B = emptyset$ ↯
The cases are exhaustive and each lead to a contradiction. $therefore,$ their union is connected. QED
For the former my proof is:
Definitions:
$$z in partial A iff exists r>0:D[z,r]cap A ne emptyset ne D[z,r]cap A^c$$
$$z in partial B iff exists r>0:D[z,r]cap B ne emptyset ne D[z,r]cap B^c$$
$$B textis closed iff partial B subseteq B iff B = B^0 cup partial B$$
Now observe for $D[z,r]cap A$:
$$D[z,r]cap A subseteq A subseteq B implies D[z,r]cap B ne emptyset$$
Thus for $D[z,r]cap A^c$:
$$D[z,r]cap A^c subseteq B iff D[z,r] subseteq B iff z in B^0$$
$$D[z,r]cap A^c subsetneq B iff D[z,r]cap B^c ne emptyset iff z in partial B$$
In either case $z in B$. QED
For the latter my proof is:
$$forall z in A=A^0, exists a>0: D[z,a] subset A =A^0 subseteq B$$
Now $$z in B^0 iff exists b>0:D[z,b] subseteq B$$
Choose $b=a$. QED
Where have I gone wrong?
real-analysis general-topology complex-analysis connectedness
asked Jul 29 at 11:34
BCLC
6,95921973
edited Aug 5 at 11:52
asked Jul 29 at 11:34
BCLC
6,95921973
asked Jul 29 at 11:34
BCLC
6,95921973
asked Jul 29 at 11:34
BCLC
6,95921973
6,95921973
add a comment |Â
add a comment |Â
2 Answers
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In any space let $F$ be a non-empty family of connected open sets such that $cap Fne emptyset.$ Let $A, B$ be disjoint open sets with $Acup B=cup F.$
For each $fin F,$ the sets $fcap A , fcap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...quad fcap Ane emptyset implies f=fcap A quad text and quad fcap Bne emptyset implies f=fcap B.$$ By contradiction, suppose $Ane emptyset ne B.$ Since $A=cup_fin F(fcap A)$ and $B=cup_fin F(fcap B),$ there exist $f_A,f_Bin F$ such that $f_Acap Ane emptyset ne f_Bcap B.$
Then by $(1)$ we have $cap Fsubset f_Acap f_B=(f_Acap A)cap (f_Bcap B)subset Acap B=emptyset,$ contrary to $emptyset ne cap F.$
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
add a comment |Â
up vote
0
down vote
Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete 0,1, f is constant.
If C is a collection of connected sets, a in $cap$C,
and f:$cup$C -> 0,1 continuous, then for all K in C,
f(K) = f(a). Whereupon, f($cup$C) = f(a).
answered Jul 29 at 22:19
William Elliot
5,0722414
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
In any space let $F$ be a non-empty family of connected open sets such that $cap Fne emptyset.$ Let $A, B$ be disjoint open sets with $Acup B=cup F.$
For each $fin F,$ the sets $fcap A , fcap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...quad fcap Ane emptyset implies f=fcap A quad text and quad fcap Bne emptyset implies f=fcap B.$$ By contradiction, suppose $Ane emptyset ne B.$ Since $A=cup_fin F(fcap A)$ and $B=cup_fin F(fcap B),$ there exist $f_A,f_Bin F$ such that $f_Acap Ane emptyset ne f_Bcap B.$
Then by $(1)$ we have $cap Fsubset f_Acap f_B=(f_Acap A)cap (f_Bcap B)subset Acap B=emptyset,$ contrary to $emptyset ne cap F.$
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
add a comment |Â
up vote
1
down vote
In any space let $F$ be a non-empty family of connected open sets such that $cap Fne emptyset.$ Let $A, B$ be disjoint open sets with $Acup B=cup F.$
For each $fin F,$ the sets $fcap A , fcap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...quad fcap Ane emptyset implies f=fcap A quad text and quad fcap Bne emptyset implies f=fcap B.$$ By contradiction, suppose $Ane emptyset ne B.$ Since $A=cup_fin F(fcap A)$ and $B=cup_fin F(fcap B),$ there exist $f_A,f_Bin F$ such that $f_Acap Ane emptyset ne f_Bcap B.$
Then by $(1)$ we have $cap Fsubset f_Acap f_B=(f_Acap A)cap (f_Bcap B)subset Acap B=emptyset,$ contrary to $emptyset ne cap F.$
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In any space let $F$ be a non-empty family of connected open sets such that $cap Fne emptyset.$ Let $A, B$ be disjoint open sets with $Acup B=cup F.$
For each $fin F,$ the sets $fcap A , fcap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...quad fcap Ane emptyset implies f=fcap A quad text and quad fcap Bne emptyset implies f=fcap B.$$ By contradiction, suppose $Ane emptyset ne B.$ Since $A=cup_fin F(fcap A)$ and $B=cup_fin F(fcap B),$ there exist $f_A,f_Bin F$ such that $f_Acap Ane emptyset ne f_Bcap B.$
Then by $(1)$ we have $cap Fsubset f_Acap f_B=(f_Acap A)cap (f_Bcap B)subset Acap B=emptyset,$ contrary to $emptyset ne cap F.$
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
In any space let $F$ be a non-empty family of connected open sets such that $cap Fne emptyset.$ Let $A, B$ be disjoint open sets with $Acup B=cup F.$
For each $fin F,$ the sets $fcap A , fcap B$ are disjoint open subsets of the connected set $f,$ and their union is $f$. So $$(1)...quad fcap Ane emptyset implies f=fcap A quad text and quad fcap Bne emptyset implies f=fcap B.$$ By contradiction, suppose $Ane emptyset ne B.$ Since $A=cup_fin F(fcap A)$ and $B=cup_fin F(fcap B),$ there exist $f_A,f_Bin F$ such that $f_Acap Ane emptyset ne f_Bcap B.$
Then by $(1)$ we have $cap Fsubset f_Acap f_B=(f_Acap A)cap (f_Bcap B)subset Acap B=emptyset,$ contrary to $emptyset ne cap F.$
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
answered Jul 30 at 18:31
DanielWainfleet
31.4k31542
31.4k31542
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
add a comment |Â
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
Thanks DanielWainfleet! I'll analyse later
– BCLC
Aug 5 at 11:53
add a comment |Â
up vote
0
down vote
Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete 0,1, f is constant.
If C is a collection of connected sets, a in $cap$C,
and f:$cup$C -> 0,1 continuous, then for all K in C,
f(K) = f(a). Whereupon, f($cup$C) = f(a).
answered Jul 29 at 22:19
William Elliot
5,0722414
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
add a comment |Â
up vote
0
down vote
Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete 0,1, f is constant.
If C is a collection of connected sets, a in $cap$C,
and f:$cup$C -> 0,1 continuous, then for all K in C,
f(K) = f(a). Whereupon, f($cup$C) = f(a).
answered Jul 29 at 22:19
William Elliot
5,0722414
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete 0,1, f is constant.
If C is a collection of connected sets, a in $cap$C,
and f:$cup$C -> 0,1 continuous, then for all K in C,
f(K) = f(a). Whereupon, f($cup$C) = f(a).
answered Jul 29 at 22:19
William Elliot
5,0722414
Any union of connected sets with a not empty intersection
is connected. Use the theorem that K is connected iff for
all continuous f:K -> discrete 0,1, f is constant.
If C is a collection of connected sets, a in $cap$C,
and f:$cup$C -> 0,1 continuous, then for all K in C,
f(K) = f(a). Whereupon, f($cup$C) = f(a).
answered Jul 29 at 22:19
William Elliot
5,0722414
answered Jul 29 at 22:19
William Elliot
5,0722414
answered Jul 29 at 22:19
William Elliot
5,0722414
answered Jul 29 at 22:19
William Elliot
5,0722414
5,0722414
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
add a comment |Â
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
'my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis'...
– BCLC
Jul 30 at 0:29
add a comment |Â
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