Analytic functions with uncountable zeros [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Suppose for an analytic function $f$ its zero set $Z_f$ is uncountable. Show that $f â¡ 0$.
Can we show that the set has a limit point? Hence the result.
complex-analysis
closed as off-topic by Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy Aug 18 at 17:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy
add a comment |Â
up vote
0
down vote
favorite
Suppose for an analytic function $f$ its zero set $Z_f$ is uncountable. Show that $f â¡ 0$.
Can we show that the set has a limit point? Hence the result.
complex-analysis
closed as off-topic by Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy Aug 18 at 17:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose for an analytic function $f$ its zero set $Z_f$ is uncountable. Show that $f â¡ 0$.
Can we show that the set has a limit point? Hence the result.
complex-analysis
Suppose for an analytic function $f$ its zero set $Z_f$ is uncountable. Show that $f â¡ 0$.
Can we show that the set has a limit point? Hence the result.
complex-analysis
asked Feb 20 '16 at 11:48
jaggu
736
736
closed as off-topic by Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy Aug 18 at 17:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy
closed as off-topic by Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy Aug 18 at 17:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â Xander Henderson, Leucippus, Strants, Brahadeesh, amWhy
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05
add a comment |Â
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
The complex plane is the union of countably many compact sets, and some of the compact sets must contain infinitely many zeros.
add a comment |Â
up vote
1
down vote
Yes every uncountable set has a limit point. More than that it has a condensation point.
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
add a comment |Â
up vote
0
down vote
The zeros of an analytic function that is not the zero function are isolated.
This means that for each zero there is a disk that contains it and no other zero. In each such disk there is a point with rational coordinates. The rational point for each disk is different because the disks are disjoint.
This gives an injection from the set of zeros into $mathbb Q times mathbb Q$, which is countable. Therefore, the set of zeros is countable.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The complex plane is the union of countably many compact sets, and some of the compact sets must contain infinitely many zeros.
add a comment |Â
up vote
4
down vote
The complex plane is the union of countably many compact sets, and some of the compact sets must contain infinitely many zeros.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The complex plane is the union of countably many compact sets, and some of the compact sets must contain infinitely many zeros.
The complex plane is the union of countably many compact sets, and some of the compact sets must contain infinitely many zeros.
answered Feb 20 '16 at 11:51
Qiyu Wen
1,360415
1,360415
add a comment |Â
add a comment |Â
up vote
1
down vote
Yes every uncountable set has a limit point. More than that it has a condensation point.
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
add a comment |Â
up vote
1
down vote
Yes every uncountable set has a limit point. More than that it has a condensation point.
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes every uncountable set has a limit point. More than that it has a condensation point.
Yes every uncountable set has a limit point. More than that it has a condensation point.
edited Aug 17 at 5:02
Cornman
2,60721128
2,60721128
answered Feb 20 '16 at 11:55
Upstart
1,680517
1,680517
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
add a comment |Â
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
I guess jaggu u dont have any of your own ideas. As we can see from your questions..please try a bit more than that i guess you're a nbhm aspirant and csir aspirant....
â Upstart
Feb 20 '16 at 12:07
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
Can you help me in doing that?
â jaggu
Feb 20 '16 at 12:34
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
The help,if any has to come from your end only..just try to give your thoughts on the questions maybe wrong but still..
â Upstart
Feb 20 '16 at 12:39
add a comment |Â
up vote
0
down vote
The zeros of an analytic function that is not the zero function are isolated.
This means that for each zero there is a disk that contains it and no other zero. In each such disk there is a point with rational coordinates. The rational point for each disk is different because the disks are disjoint.
This gives an injection from the set of zeros into $mathbb Q times mathbb Q$, which is countable. Therefore, the set of zeros is countable.
add a comment |Â
up vote
0
down vote
The zeros of an analytic function that is not the zero function are isolated.
This means that for each zero there is a disk that contains it and no other zero. In each such disk there is a point with rational coordinates. The rational point for each disk is different because the disks are disjoint.
This gives an injection from the set of zeros into $mathbb Q times mathbb Q$, which is countable. Therefore, the set of zeros is countable.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The zeros of an analytic function that is not the zero function are isolated.
This means that for each zero there is a disk that contains it and no other zero. In each such disk there is a point with rational coordinates. The rational point for each disk is different because the disks are disjoint.
This gives an injection from the set of zeros into $mathbb Q times mathbb Q$, which is countable. Therefore, the set of zeros is countable.
The zeros of an analytic function that is not the zero function are isolated.
This means that for each zero there is a disk that contains it and no other zero. In each such disk there is a point with rational coordinates. The rational point for each disk is different because the disks are disjoint.
This gives an injection from the set of zeros into $mathbb Q times mathbb Q$, which is countable. Therefore, the set of zeros is countable.
answered Feb 20 '16 at 12:42
lhf
156k9161367
156k9161367
add a comment |Â
add a comment |Â
This is false without further hypotheses. Start by telling us what the domain of $f$ is.
â zhw.
Aug 17 at 6:05