the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1text{ and }y in mathbb{N}}$ is which of the following statement...











up vote
-2
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In $mathbb{R}^2$ with usual topology, the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1 text{ and } y in mathbb{N}}$ is
which of the following statement is True ?



a) neither closed nor bounded



b)closed but not bounded



c)bounded but not closed



d)closed and bounded



I thinks it will be bounded but not closed because $(x,-y)$ is an open set that is option c) will be true



Is its True ?



any hints/solution will apprecaited



thanks u






general-topology







share|cite|improve this question















closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 at 9:08


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." – José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
    – José Carlos Santos
    Nov 23 at 18:24










  • @JoséCarlosSantos any Hints ??
    – santosh
    Nov 23 at 18:33






  • 1




    The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
    – Mihail
    Nov 23 at 18:59










  • ya u r right im trying my best next time @Mihail
    – santosh
    Nov 23 at 19:05















up vote
-2
down vote

favorite












In $mathbb{R}^2$ with usual topology, the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1 text{ and } y in mathbb{N}}$ is
which of the following statement is True ?



a) neither closed nor bounded



b)closed but not bounded



c)bounded but not closed



d)closed and bounded



I thinks it will be bounded but not closed because $(x,-y)$ is an open set that is option c) will be true



Is its True ?



any hints/solution will apprecaited



thanks u






general-topology







share|cite|improve this question















closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 at 9:08


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." – José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 2




    What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
    – José Carlos Santos
    Nov 23 at 18:24










  • @JoséCarlosSantos any Hints ??
    – santosh
    Nov 23 at 18:33






  • 1




    The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
    – Mihail
    Nov 23 at 18:59










  • ya u r right im trying my best next time @Mihail
    – santosh
    Nov 23 at 19:05













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











In $mathbb{R}^2$ with usual topology, the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1 text{ and } y in mathbb{N}}$ is
which of the following statement is True ?



a) neither closed nor bounded



b)closed but not bounded



c)bounded but not closed



d)closed and bounded



I thinks it will be bounded but not closed because $(x,-y)$ is an open set that is option c) will be true



Is its True ?



any hints/solution will apprecaited



thanks u






general-topology







share|cite|improve this question















In $mathbb{R}^2$ with usual topology, the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1 text{ and } y in mathbb{N}}$ is
which of the following statement is True ?



a) neither closed nor bounded



b)closed but not bounded



c)bounded but not closed



d)closed and bounded



I thinks it will be bounded but not closed because $(x,-y)$ is an open set that is option c) will be true



Is its True ?



any hints/solution will apprecaited



thanks u






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 20:01









Lord Shark the Unknown

100k958131




100k958131










asked Nov 23 at 18:18









santosh

969




969




closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 at 9:08


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." – José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 at 9:08


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." – José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
    – José Carlos Santos
    Nov 23 at 18:24










  • @JoséCarlosSantos any Hints ??
    – santosh
    Nov 23 at 18:33






  • 1




    The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
    – Mihail
    Nov 23 at 18:59










  • ya u r right im trying my best next time @Mihail
    – santosh
    Nov 23 at 19:05














  • 2




    What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
    – José Carlos Santos
    Nov 23 at 18:24










  • @JoséCarlosSantos any Hints ??
    – santosh
    Nov 23 at 18:33






  • 1




    The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
    – Mihail
    Nov 23 at 18:59










  • ya u r right im trying my best next time @Mihail
    – santosh
    Nov 23 at 19:05








2




2




What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
– José Carlos Santos
Nov 23 at 18:24




What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
– José Carlos Santos
Nov 23 at 18:24












@JoséCarlosSantos any Hints ??
– santosh
Nov 23 at 18:33




@JoséCarlosSantos any Hints ??
– santosh
Nov 23 at 18:33




1




1




The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
– Mihail
Nov 23 at 18:59




The way you're asking this question made me check several of your previous questions on mathstack. It looks like you just post the problems without making any attempts of acquainting yourself with the definitions and showing what have you tried. I'm sorry to sound rude but it's not the way math is learned.
– Mihail
Nov 23 at 18:59












ya u r right im trying my best next time @Mihail
– santosh
Nov 23 at 19:05




ya u r right im trying my best next time @Mihail
– santosh
Nov 23 at 19:05










1 Answer
1






active

oldest

votes

















up vote
-1
down vote



accepted










Surely it is not bounded since $(0,-1)$ , $(0,-2)$ , $(0,-3)$ , $(0,-4)$ , ... belong to the set. Also the set is closed since the minimum distance between any two distinct members of $U$ is at least 1.






share|cite|improve this answer





















  • ,,,how minimum distance 1 implies closed set ?? im not getting this ???
    – santosh
    Nov 23 at 18:47






  • 1




    Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
    – Mostafa Ayaz
    Nov 23 at 18:48






  • 1




    In fact after a while the terms of the convergent sequence become equal to their limit
    – Mostafa Ayaz
    Nov 23 at 18:49










  • thanks u @Mostafa
    – santosh
    Nov 23 at 18:54






  • 1




    You're welcome. Hope it help...
    – Mostafa Ayaz
    Nov 23 at 18:55


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
-1
down vote



accepted










Surely it is not bounded since $(0,-1)$ , $(0,-2)$ , $(0,-3)$ , $(0,-4)$ , ... belong to the set. Also the set is closed since the minimum distance between any two distinct members of $U$ is at least 1.






share|cite|improve this answer





















  • ,,,how minimum distance 1 implies closed set ?? im not getting this ???
    – santosh
    Nov 23 at 18:47






  • 1




    Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
    – Mostafa Ayaz
    Nov 23 at 18:48






  • 1




    In fact after a while the terms of the convergent sequence become equal to their limit
    – Mostafa Ayaz
    Nov 23 at 18:49










  • thanks u @Mostafa
    – santosh
    Nov 23 at 18:54






  • 1




    You're welcome. Hope it help...
    – Mostafa Ayaz
    Nov 23 at 18:55















up vote
-1
down vote



accepted










Surely it is not bounded since $(0,-1)$ , $(0,-2)$ , $(0,-3)$ , $(0,-4)$ , ... belong to the set. Also the set is closed since the minimum distance between any two distinct members of $U$ is at least 1.






share|cite|improve this answer





















  • ,,,how minimum distance 1 implies closed set ?? im not getting this ???
    – santosh
    Nov 23 at 18:47






  • 1




    Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
    – Mostafa Ayaz
    Nov 23 at 18:48






  • 1




    In fact after a while the terms of the convergent sequence become equal to their limit
    – Mostafa Ayaz
    Nov 23 at 18:49










  • thanks u @Mostafa
    – santosh
    Nov 23 at 18:54






  • 1




    You're welcome. Hope it help...
    – Mostafa Ayaz
    Nov 23 at 18:55













up vote
-1
down vote



accepted







up vote
-1
down vote



accepted






Surely it is not bounded since $(0,-1)$ , $(0,-2)$ , $(0,-3)$ , $(0,-4)$ , ... belong to the set. Also the set is closed since the minimum distance between any two distinct members of $U$ is at least 1.






share|cite|improve this answer












Surely it is not bounded since $(0,-1)$ , $(0,-2)$ , $(0,-3)$ , $(0,-4)$ , ... belong to the set. Also the set is closed since the minimum distance between any two distinct members of $U$ is at least 1.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 at 18:44









Mostafa Ayaz

13.6k3836




13.6k3836












  • ,,,how minimum distance 1 implies closed set ?? im not getting this ???
    – santosh
    Nov 23 at 18:47






  • 1




    Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
    – Mostafa Ayaz
    Nov 23 at 18:48






  • 1




    In fact after a while the terms of the convergent sequence become equal to their limit
    – Mostafa Ayaz
    Nov 23 at 18:49










  • thanks u @Mostafa
    – santosh
    Nov 23 at 18:54






  • 1




    You're welcome. Hope it help...
    – Mostafa Ayaz
    Nov 23 at 18:55


















  • ,,,how minimum distance 1 implies closed set ?? im not getting this ???
    – santosh
    Nov 23 at 18:47






  • 1




    Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
    – Mostafa Ayaz
    Nov 23 at 18:48






  • 1




    In fact after a while the terms of the convergent sequence become equal to their limit
    – Mostafa Ayaz
    Nov 23 at 18:49










  • thanks u @Mostafa
    – santosh
    Nov 23 at 18:54






  • 1




    You're welcome. Hope it help...
    – Mostafa Ayaz
    Nov 23 at 18:55
















,,,how minimum distance 1 implies closed set ?? im not getting this ???
– santosh
Nov 23 at 18:47




,,,how minimum distance 1 implies closed set ?? im not getting this ???
– santosh
Nov 23 at 18:47




1




1




Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
– Mostafa Ayaz
Nov 23 at 18:48




Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
– Mostafa Ayaz
Nov 23 at 18:48




1




1




In fact after a while the terms of the convergent sequence become equal to their limit
– Mostafa Ayaz
Nov 23 at 18:49




In fact after a while the terms of the convergent sequence become equal to their limit
– Mostafa Ayaz
Nov 23 at 18:49












thanks u @Mostafa
– santosh
Nov 23 at 18:54




thanks u @Mostafa
– santosh
Nov 23 at 18:54




1




1




You're welcome. Hope it help...
– Mostafa Ayaz
Nov 23 at 18:55




You're welcome. Hope it help...
– Mostafa Ayaz
Nov 23 at 18:55



-

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