Open or Closed set given a region in complex plane [closed]
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Is the region given by 1) Re(z)<=-1 2) 0 an open set , closed set or, neither open nor closed set?
I think 1) should be closed since it contains all boundary points
2) im not sure how I should consider z!=0
complex-analysis
asked Nov 14 at 8:55
user856707
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closed as off-topic by 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy Nov 14 at 11:07
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." – 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
Is the region given by 1) Re(z)<=-1 2) 0 an open set , closed set or, neither open nor closed set?
I think 1) should be closed since it contains all boundary points
2) im not sure how I should consider z!=0
complex-analysis
asked Nov 14 at 8:55
user856707
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy Nov 14 at 11:07
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." – 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Is the region given by 1) Re(z)<=-1 2) 0 an open set , closed set or, neither open nor closed set?
I think 1) should be closed since it contains all boundary points
2) im not sure how I should consider z!=0
complex-analysis
asked Nov 14 at 8:55
user856707
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Is the region given by 1) Re(z)<=-1 2) 0 an open set , closed set or, neither open nor closed set?
I think 1) should be closed since it contains all boundary points
2) im not sure how I should consider z!=0
complex-analysis
complex-analysis
asked Nov 14 at 8:55
user856707
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 14 at 8:55
user856707
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 14 at 9:08
asked Nov 14 at 8:55
user856707
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 14 at 8:55
user856707
11
asked Nov 14 at 8:55
user856707
11
11
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user856707 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy Nov 14 at 11:07
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." – 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy Nov 14 at 11:07
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." – 5xum, José Carlos Santos, ArsenBerk, Christopher, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58
add a comment |
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58
add a comment |
1 Answer
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Let $R={z in mathbb C: Re(z) le 1}$. If $(z_n)$ is a convergent sequence in $R$ with limit $z_0$, then $Re(z_n) to Re(z_0)$. Since $Re(z_n) le -1$ for all $n$, we get $Re(z_0) le -1$. Thus $z_0 in R$.
Conclusion: $R$ is closed.
answered Nov 14 at 10:03
Fred
41.9k1642
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $R={z in mathbb C: Re(z) le 1}$. If $(z_n)$ is a convergent sequence in $R$ with limit $z_0$, then $Re(z_n) to Re(z_0)$. Since $Re(z_n) le -1$ for all $n$, we get $Re(z_0) le -1$. Thus $z_0 in R$.
Conclusion: $R$ is closed.
answered Nov 14 at 10:03
Fred
41.9k1642
add a comment |
up vote
0
down vote
Let $R={z in mathbb C: Re(z) le 1}$. If $(z_n)$ is a convergent sequence in $R$ with limit $z_0$, then $Re(z_n) to Re(z_0)$. Since $Re(z_n) le -1$ for all $n$, we get $Re(z_0) le -1$. Thus $z_0 in R$.
Conclusion: $R$ is closed.
answered Nov 14 at 10:03
Fred
41.9k1642
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $R={z in mathbb C: Re(z) le 1}$. If $(z_n)$ is a convergent sequence in $R$ with limit $z_0$, then $Re(z_n) to Re(z_0)$. Since $Re(z_n) le -1$ for all $n$, we get $Re(z_0) le -1$. Thus $z_0 in R$.
Conclusion: $R$ is closed.
answered Nov 14 at 10:03
Fred
41.9k1642
Let $R={z in mathbb C: Re(z) le 1}$. If $(z_n)$ is a convergent sequence in $R$ with limit $z_0$, then $Re(z_n) to Re(z_0)$. Since $Re(z_n) le -1$ for all $n$, we get $Re(z_0) le -1$. Thus $z_0 in R$.
Conclusion: $R$ is closed.
answered Nov 14 at 10:03
Fred
41.9k1642
answered Nov 14 at 10:03
Fred
41.9k1642
answered Nov 14 at 10:03
Fred
41.9k1642
answered Nov 14 at 10:03
Fred
41.9k1642
41.9k1642
add a comment |
add a comment |
Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? Don't worry if it's wrong - that's what we're here for.
– 5xum
Nov 14 at 8:58
Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote.
– 5xum
Nov 14 at 8:58