the set $U ={(x,-y) in mathbb{R}^2 : x=0,1,-1text{ and }y in mathbb{N}}$ is which of the following statement...
-2
$begingroup$
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
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
$endgroup$
closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 '18 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.
add a comment |
-2
$begingroup$
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
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
$endgroup$
closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 '18 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
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05
add a comment |
-2
-2
-2
$begingroup$
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
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
$endgroup$
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
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
edited Nov 23 '18 at 20:01
Lord Shark the Unknown
102k959132
102k959132
asked Nov 23 '18 at 18:18
santoshsantosh
1019
asked Nov 23 '18 at 18:18
santoshsantosh
1019
asked Nov 23 '18 at 18:18
santoshsantosh
1019
1019
closed as off-topic by José Carlos Santos, Leucippus, Cesareo, Brahadeesh, Rebellos Nov 24 '18 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 '18 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
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05
add a comment |
2
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05
2
2
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
1
1
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05
add a comment |
1 Answer
1
active
oldest
votes
-1
$begingroup$
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.
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
-1
$begingroup$
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.
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
add a comment |
-1
$begingroup$
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.
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
add a comment |
-1
-1
-1
$begingroup$
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.
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
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.
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 18:44
Mostafa AyazMostafa Ayaz
14.7k3938
14.7k3938
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
add a comment |
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
$begingroup$
,,,how minimum distance 1 implies closed set ?? im not getting this ???
$endgroup$
– santosh
Nov 23 '18 at 18:47
1
1
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
$begingroup$
Then if the terms of a sequence fall near their limit closer than the distance of 1 they must urgently be the limit itself
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:48
1
1
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
In fact after a while the terms of the convergent sequence become equal to their limit
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:49
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
$begingroup$
thanks u @Mostafa
$endgroup$
– santosh
Nov 23 '18 at 18:54
1
1
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
$begingroup$
You're welcome. Hope it help...
$endgroup$
– Mostafa Ayaz
Nov 23 '18 at 18:55
add a comment |
2
$begingroup$
What does “$(x,-y)$ is an open set mean”? After all, $(x,-y)$ is a point not a set.
$endgroup$
– José Carlos Santos
Nov 23 '18 at 18:24
$begingroup$
@JoséCarlosSantos any Hints ??
$endgroup$
– santosh
Nov 23 '18 at 18:33
1
$begingroup$
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.
$endgroup$
– Mihail
Nov 23 '18 at 18:59
$begingroup$
ya u r right im trying my best next time @Mihail
$endgroup$
– santosh
Nov 23 '18 at 19:05