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Compactness of a set in the complex plane.
Clash Royale CLAN TAG#URR8PPP
up vote
8
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Consider $$A=Big(z_1,z_2) in BbbC^2 : z_1^2+z_2^2=1Big$$
Is $A$ compact?
My try: If $A$ is compact, then it is closed and bounded
But it is not bounded! Since $(n,sqrt1-n^2);nin BbbN$ satisfies this and $BbbN$ is unbounded. So not compact.
Am I right? If not, any help?
complex-analysis complex-geometry
edited Sep 6 at 6:14
Filippo De Bortoli
997521
asked Sep 6 at 5:55
LDM
730314
add a comment |Â
up vote
8
down vote
favorite
Consider $$A=Big(z_1,z_2) in BbbC^2 : z_1^2+z_2^2=1Big$$
Is $A$ compact?
My try: If $A$ is compact, then it is closed and bounded
But it is not bounded! Since $(n,sqrt1-n^2);nin BbbN$ satisfies this and $BbbN$ is unbounded. So not compact.
Am I right? If not, any help?
complex-analysis complex-geometry
edited Sep 6 at 6:14
Filippo De Bortoli
997521
asked Sep 6 at 5:55
LDM
730314
3
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
1
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Consider $$A=Big(z_1,z_2) in BbbC^2 : z_1^2+z_2^2=1Big$$
Is $A$ compact?
My try: If $A$ is compact, then it is closed and bounded
But it is not bounded! Since $(n,sqrt1-n^2);nin BbbN$ satisfies this and $BbbN$ is unbounded. So not compact.
Am I right? If not, any help?
complex-analysis complex-geometry
edited Sep 6 at 6:14
Filippo De Bortoli
997521
asked Sep 6 at 5:55
LDM
730314
Consider $$A=Big(z_1,z_2) in BbbC^2 : z_1^2+z_2^2=1Big$$
Is $A$ compact?
My try: If $A$ is compact, then it is closed and bounded
But it is not bounded! Since $(n,sqrt1-n^2);nin BbbN$ satisfies this and $BbbN$ is unbounded. So not compact.
Am I right? If not, any help?
complex-analysis complex-geometry
edited Sep 6 at 6:14
Filippo De Bortoli
997521
asked Sep 6 at 5:55
LDM
730314
edited Sep 6 at 6:14
Filippo De Bortoli
997521
edited Sep 6 at 6:14
Filippo De Bortoli
997521
edited Sep 6 at 6:14
Filippo De Bortoli
997521
997521
asked Sep 6 at 5:55
LDM
730314
asked Sep 6 at 5:55
LDM
730314
asked Sep 6 at 5:55
LDM
730314
730314
3
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
1
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59
add a comment |Â
3
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
1
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59
3
3
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
1
1
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, isqrtn^2-1)$ instead. However, not everyone are as opposed to writing $sqrt-1$ as I am.
Apart from that it looks good. Well done!
answered Sep 6 at 6:05
Arthur
101k795176
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, isqrtn^2-1)$ instead. However, not everyone are as opposed to writing $sqrt-1$ as I am.
Apart from that it looks good. Well done!
answered Sep 6 at 6:05
Arthur
101k795176
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
add a comment |Â
up vote
8
down vote
accepted
I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, isqrtn^2-1)$ instead. However, not everyone are as opposed to writing $sqrt-1$ as I am.
Apart from that it looks good. Well done!
answered Sep 6 at 6:05
Arthur
101k795176
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, isqrtn^2-1)$ instead. However, not everyone are as opposed to writing $sqrt-1$ as I am.
Apart from that it looks good. Well done!
answered Sep 6 at 6:05
Arthur
101k795176
I personally dislike the use of the square root sign in this context (it stops being well-defined the moment you start using it on negative numbers), but that may be fixed by writing, say, $(n, isqrtn^2-1)$ instead. However, not everyone are as opposed to writing $sqrt-1$ as I am.
Apart from that it looks good. Well done!
answered Sep 6 at 6:05
Arthur
101k795176
edited Sep 6 at 6:08
answered Sep 6 at 6:05
Arthur
101k795176
answered Sep 6 at 6:05
Arthur
101k795176
answered Sep 6 at 6:05
Arthur
101k795176
101k795176
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
add a comment |Â
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
1
1
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
@LDM Then you (and probably your teacher and / or textbook) are not as opposed to the idea as I am. I think it is not well-defined but I realize there are other conventions. Just be advised that rules you have learned, like $sqrtab=sqrt asqrt b$, stops working in that case.
– Arthur
Sep 6 at 6:10
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Ok! I understand! Thank you sir
– LDM
Sep 6 at 6:11
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
Good remark.( +1)
– Math_QED
Sep 6 at 7:29
add a comment |Â
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3
In a competition, this question would be confusing since the corresponding equation in $mathbb R^2$ corresponds to a compact set.You are right, this set is unbounded, hence definitely not compact.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Sep 6 at 5:57
1
complex analysis tag may be needed instead.
– mrs
Sep 6 at 5:59