Mixing
Sketch the sets in $mathbb C$ and determine whether they are open, closed, or neither; bounded; connected and their boundaries. [closed]
Clash Royale CLAN TAG#URR8PPP
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A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.27,8
Here are my answers:
- Where have I gone wrong?
- About connected, my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis, how exactly do we prove those sets are connected? Suppose they're not connected and then derive a contradiction like in my other question Elementary topology of $mathbb C$: Union of 2 regions with nonempty intersection is a region ?
- Is there a way to check this using computers? Not sure I can use Wolfram Alpha or guess I don't know how. Are there programs for this? Like can I do this in Matlab, scilab or R?
general-topology complex-analysis convex-analysis connectedness path-connected
asked Jul 30 at 10:36
BCLC
6,99221973
closed as too broad by John Ma, Lord Shark the Unknown, max_zorn, Jose Arnaldo Bebita Dris, Mostafa Ayaz Aug 2 at 9:36
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.27,8
Here are my answers:
- Where have I gone wrong?
- About connected, my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis, how exactly do we prove those sets are connected? Suppose they're not connected and then derive a contradiction like in my other question Elementary topology of $mathbb C$: Union of 2 regions with nonempty intersection is a region ?
- Is there a way to check this using computers? Not sure I can use Wolfram Alpha or guess I don't know how. Are there programs for this? Like can I do this in Matlab, scilab or R?
general-topology complex-analysis convex-analysis connectedness path-connected
asked Jul 30 at 10:36
BCLC
6,99221973
closed as too broad by John Ma, Lord Shark the Unknown, max_zorn, Jose Arnaldo Bebita Dris, Mostafa Ayaz Aug 2 at 9:36
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
1
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
1
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
1
Why down vote…?
– xbh
Jul 31 at 2:47
 |Â
show 1 more comment
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.27,8
Here are my answers:
- Where have I gone wrong?
- About connected, my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis, how exactly do we prove those sets are connected? Suppose they're not connected and then derive a contradiction like in my other question Elementary topology of $mathbb C$: Union of 2 regions with nonempty intersection is a region ?
- Is there a way to check this using computers? Not sure I can use Wolfram Alpha or guess I don't know how. Are there programs for this? Like can I do this in Matlab, scilab or R?
general-topology complex-analysis convex-analysis connectedness path-connected
asked Jul 30 at 10:36
BCLC
6,99221973
A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 1.27,8
Here are my answers:
- Where have I gone wrong?
- About connected, my topology is so far limited to the elementary topology in elementary analysis, complex analysis and real analysis, how exactly do we prove those sets are connected? Suppose they're not connected and then derive a contradiction like in my other question Elementary topology of $mathbb C$: Union of 2 regions with nonempty intersection is a region ?
- Is there a way to check this using computers? Not sure I can use Wolfram Alpha or guess I don't know how. Are there programs for this? Like can I do this in Matlab, scilab or R?
general-topology complex-analysis convex-analysis connectedness path-connected
asked Jul 30 at 10:36
BCLC
6,99221973
edited Aug 5 at 10:59
asked Jul 30 at 10:36
BCLC
6,99221973
asked Jul 30 at 10:36
BCLC
6,99221973
asked Jul 30 at 10:36
BCLC
6,99221973
6,99221973
closed as too broad by John Ma, Lord Shark the Unknown, max_zorn, Jose Arnaldo Bebita Dris, Mostafa Ayaz Aug 2 at 9:36
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by John Ma, Lord Shark the Unknown, max_zorn, Jose Arnaldo Bebita Dris, Mostafa Ayaz Aug 2 at 9:36
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
1
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
1
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
1
Why down vote…?
– xbh
Jul 31 at 2:47
 |Â
show 1 more comment
3
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
1
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
1
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
1
Why down vote…?
– xbh
Jul 31 at 2:47
3
3
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
1
1
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
1
1
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
1
1
Why down vote…?
– xbh
Jul 31 at 2:47
Why down vote…?
– xbh
Jul 31 at 2:47
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
a) Wrong. It's the open disk $D(-3,2)$. In particular, it is open and it is not closed.
b) Right.
c) Right, but you should write $1$ instead of $(1,0)$.
d) Right, except that it is closed.
e) Right.
f) Right.
For each set, the easiest way to prove that it is connected consists in proving that it is path-connected.
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
 |Â
show 3 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) Wrong. It's the open disk $D(-3,2)$. In particular, it is open and it is not closed.
b) Right.
c) Right, but you should write $1$ instead of $(1,0)$.
d) Right, except that it is closed.
e) Right.
f) Right.
For each set, the easiest way to prove that it is connected consists in proving that it is path-connected.
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
 |Â
show 3 more comments
up vote
1
down vote
accepted
a) Wrong. It's the open disk $D(-3,2)$. In particular, it is open and it is not closed.
b) Right.
c) Right, but you should write $1$ instead of $(1,0)$.
d) Right, except that it is closed.
e) Right.
f) Right.
For each set, the easiest way to prove that it is connected consists in proving that it is path-connected.
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
a) Wrong. It's the open disk $D(-3,2)$. In particular, it is open and it is not closed.
b) Right.
c) Right, but you should write $1$ instead of $(1,0)$.
d) Right, except that it is closed.
e) Right.
f) Right.
For each set, the easiest way to prove that it is connected consists in proving that it is path-connected.
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
a) Wrong. It's the open disk $D(-3,2)$. In particular, it is open and it is not closed.
b) Right.
c) Right, but you should write $1$ instead of $(1,0)$.
d) Right, except that it is closed.
e) Right.
f) Right.
For each set, the easiest way to prove that it is connected consists in proving that it is path-connected.
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
answered Jul 30 at 10:45
José Carlos Santos
112k1696172
112k1696172
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
 |Â
show 3 more comments
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Thanks, José Carlos Santos! Re(a): No idea why I thought closed disc. Been weeks since I did this and am just revising now. I get that it's open. Re connected: Oh like in Thm 1.12 ('If any two points in G ⊆ C can be connected by a path in G, then G is connected')? Re computers: Any suggestions please?
– BCLC
Jul 30 at 12:17
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
Re path connected: How would you go about proving they are path connected please? For a,b,d,e: I guess we can say for path connected that we can draw a line between any 2 points s.t. the line is still in the set ('convex' isn't introduced in Ch1). What about c and f? Hmmm...perhaps respectively semi-circle and parabola instead of line if line fails?
– BCLC
Jul 30 at 12:30
2
2
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
@BCLC Yeah, that's the gist. For c) try an arc and a line segment. For f) use parabola and a line segment.
– xbh
Jul 30 at 13:25
1
1
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
@BCLC Sorry for my non-rigorous expression. I mean a line segment.
– xbh
Jul 31 at 1:45
1
1
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
@BCLC Right. An arc of a circle along with a line segment always works in c). Similar thing for f).
– xbh
Jul 31 at 2:47
 |Â
show 3 more comments
3
Why do you think (a) describes a closed disk?
– xbh
Jul 30 at 10:39
@xbh Re(a): No idea. Been weeks since I did this and am just revising now. I get that it's open. Thanks!
– BCLC
Jul 30 at 12:16
1
Glad to help. Maybe you just get exhausted and did not notice this. It's fine. For mathematical softwares, WolframAlpha is applicable [like drawing the graph of sets], but those properties needs to be verified by you.
– xbh
Jul 30 at 12:32
1
And yes, path connectedness implies connectedness. The former one seems much more acceptable from our intuition.
– xbh
Jul 30 at 12:34
1
Why down vote…?
– xbh
Jul 31 at 2:47