Given set U is first countable or not?
up vote
-1
down vote
favorite
In $mathbb{R}$ with usual topology ,the set $U ={ x in mathbb{R} : -1le x le 1 , ,x neq 0}$ is
Choose the correct statement
$a)$ Neither hausdorff nor First counatble
$b)$ Hausdorff
$c)$ First countable
$d)$both hausdorff and first countable
My attempt :set $U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint set, From this i can concnclude that $U$ is hausdorff
Im confusing that it is First countable or not ?
Any hints/solution will be appreciated
thanks u
general-topology separation-axioms first-countable
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
add a comment |
up vote
-1
down vote
favorite
In $mathbb{R}$ with usual topology ,the set $U ={ x in mathbb{R} : -1le x le 1 , ,x neq 0}$ is
Choose the correct statement
$a)$ Neither hausdorff nor First counatble
$b)$ Hausdorff
$c)$ First countable
$d)$both hausdorff and first countable
My attempt :set $U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint set, From this i can concnclude that $U$ is hausdorff
Im confusing that it is First countable or not ?
Any hints/solution will be appreciated
thanks u
general-topology separation-axioms first-countable
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
In $mathbb{R}$ with usual topology ,the set $U ={ x in mathbb{R} : -1le x le 1 , ,x neq 0}$ is
Choose the correct statement
$a)$ Neither hausdorff nor First counatble
$b)$ Hausdorff
$c)$ First countable
$d)$both hausdorff and first countable
My attempt :set $U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint set, From this i can concnclude that $U$ is hausdorff
Im confusing that it is First countable or not ?
Any hints/solution will be appreciated
thanks u
general-topology separation-axioms first-countable
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
In $mathbb{R}$ with usual topology ,the set $U ={ x in mathbb{R} : -1le x le 1 , ,x neq 0}$ is
Choose the correct statement
$a)$ Neither hausdorff nor First counatble
$b)$ Hausdorff
$c)$ First countable
$d)$both hausdorff and first countable
My attempt :set $U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint set, From this i can concnclude that $U$ is hausdorff
Im confusing that it is First countable or not ?
Any hints/solution will be appreciated
thanks u
general-topology separation-axioms first-countable
general-topology separation-axioms first-countable
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
edited Nov 22 at 13:43
José Carlos Santos
146k22117217
146k22117217
asked Nov 22 at 13:39
jasmine
1,505416
asked Nov 22 at 13:39
jasmine
1,505416
asked Nov 22 at 13:39
jasmine
1,505416
1,505416
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
The usual topology is induced by a metric and every metric space is first-countable.
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
add a comment |
up vote
2
down vote
$mathbb{R}$ with usual topology is also a metric space. So $mathbb{R}$ is first countable. Hence any subspace is also first countable.
answered Nov 22 at 13:43
Offlaw
2649
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
add a comment |
up vote
1
down vote
$U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint
set, From this i can concnclude that $U$ is hausdorff
The fact that $U$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.
For the first countable property, google is your friend.
answered Nov 22 at 13:42
5xum
89.4k393161
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The usual topology is induced by a metric and every metric space is first-countable.
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
add a comment |
up vote
1
down vote
accepted
The usual topology is induced by a metric and every metric space is first-countable.
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The usual topology is induced by a metric and every metric space is first-countable.
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
The usual topology is induced by a metric and every metric space is first-countable.
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
answered Nov 22 at 13:41
José Carlos Santos
146k22117217
146k22117217
add a comment |
add a comment |
up vote
2
down vote
$mathbb{R}$ with usual topology is also a metric space. So $mathbb{R}$ is first countable. Hence any subspace is also first countable.
answered Nov 22 at 13:43
Offlaw
2649
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
add a comment |
up vote
2
down vote
$mathbb{R}$ with usual topology is also a metric space. So $mathbb{R}$ is first countable. Hence any subspace is also first countable.
answered Nov 22 at 13:43
Offlaw
2649
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
add a comment |
up vote
2
down vote
up vote
2
down vote
$mathbb{R}$ with usual topology is also a metric space. So $mathbb{R}$ is first countable. Hence any subspace is also first countable.
answered Nov 22 at 13:43
Offlaw
2649
$mathbb{R}$ with usual topology is also a metric space. So $mathbb{R}$ is first countable. Hence any subspace is also first countable.
answered Nov 22 at 13:43
Offlaw
2649
answered Nov 22 at 13:43
Offlaw
2649
answered Nov 22 at 13:43
Offlaw
2649
answered Nov 22 at 13:43
Offlaw
2649
2649
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
add a comment |
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
1
1
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
To check Hausdorffness use definition.
– Offlaw
Nov 22 at 13:47
add a comment |
up vote
1
down vote
$U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint
set, From this i can concnclude that $U$ is hausdorff
The fact that $U$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.
For the first countable property, google is your friend.
answered Nov 22 at 13:42
5xum
89.4k393161
add a comment |
up vote
1
down vote
$U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint
set, From this i can concnclude that $U$ is hausdorff
The fact that $U$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.
For the first countable property, google is your friend.
answered Nov 22 at 13:42
5xum
89.4k393161
add a comment |
up vote
1
down vote
up vote
1
down vote
$U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint
set, From this i can concnclude that $U$ is hausdorff
The fact that $U$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.
For the first countable property, google is your friend.
answered Nov 22 at 13:42
5xum
89.4k393161
$U$ can be written as $[-1,0)$ and $(0,1]$ which are two disjoint
set, From this i can concnclude that $U$ is hausdorff
The fact that $U$ can be written as a union of two disjoint sets has nothing to do with the set being Hausforff or not.
For the first countable property, google is your friend.
answered Nov 22 at 13:42
5xum
89.4k393161
answered Nov 22 at 13:42
5xum
89.4k393161
answered Nov 22 at 13:42
5xum
89.4k393161
answered Nov 22 at 13:42
5xum
89.4k393161
89.4k393161
add a comment |
add a comment |
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