some confusion about open subset?
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Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.
I have some confusion in my mind that is
Is $K$ is an open subset of $mathbb{Q}$ ?
My attempt : my answer is No,
$K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.
From this I can conclude that K is not open subset of $mathbb{Q}$
Is it True ?
general-topology proof-verification compactness
edited 31 mins ago
José Carlos Santos
146k22116216
asked 38 mins ago
jasmine
1,448416
add a comment |
up vote
2
down vote
favorite
Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.
I have some confusion in my mind that is
Is $K$ is an open subset of $mathbb{Q}$ ?
My attempt : my answer is No,
$K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.
From this I can conclude that K is not open subset of $mathbb{Q}$
Is it True ?
general-topology proof-verification compactness
edited 31 mins ago
José Carlos Santos
146k22116216
asked 38 mins ago
jasmine
1,448416
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.
I have some confusion in my mind that is
Is $K$ is an open subset of $mathbb{Q}$ ?
My attempt : my answer is No,
$K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.
From this I can conclude that K is not open subset of $mathbb{Q}$
Is it True ?
general-topology proof-verification compactness
edited 31 mins ago
José Carlos Santos
146k22116216
asked 38 mins ago
jasmine
1,448416
Consider the set of rational number $mathbb{Q}$ as a subset of $mathbb{R}$ with the usual metric. Let $K = [ sqrt 2, sqrt 3] cap mathbb{Q}$.
I have some confusion in my mind that is
Is $K$ is an open subset of $mathbb{Q}$ ?
My attempt : my answer is No,
$K=[sqrt 2, sqrt 3]cap Bbb{Q}={q in Bbb{Q}|sqrt 2< q< sqrt 3}$ where$[sqrt 2, sqrt 3]$ is closed in $Bbb{R}$.
From this I can conclude that K is not open subset of $mathbb{Q}$
Is it True ?
general-topology proof-verification compactness
general-topology proof-verification compactness
edited 31 mins ago
José Carlos Santos
146k22116216
asked 38 mins ago
jasmine
1,448416
edited 31 mins ago
José Carlos Santos
146k22116216
asked 38 mins ago
jasmine
1,448416
edited 31 mins ago
José Carlos Santos
146k22116216
edited 31 mins ago
José Carlos Santos
146k22116216
edited 31 mins ago
José Carlos Santos
146k22116216
146k22116216
asked 38 mins ago
jasmine
1,448416
asked 38 mins ago
jasmine
1,448416
asked 38 mins ago
jasmine
1,448416
1,448416
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.
answered 36 mins ago
José Carlos Santos
146k22116216
add a comment |
up vote
7
down vote
No that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.
Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
answered 36 mins ago
Yanko
5,432722
add a comment |
up vote
1
down vote
A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.
Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.
answered 36 mins ago
Arthur
110k7104186
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
3
down vote
accepted
Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.
answered 36 mins ago
José Carlos Santos
146k22116216
add a comment |
up vote
3
down vote
accepted
Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.
answered 36 mins ago
José Carlos Santos
146k22116216
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.
answered 36 mins ago
José Carlos Santos
146k22116216
Yes, $K$ is an open subset of $mathbb Q$, since $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ is an open subset of $mathbb R$.
answered 36 mins ago
José Carlos Santos
146k22116216
answered 36 mins ago
José Carlos Santos
146k22116216
answered 36 mins ago
José Carlos Santos
146k22116216
answered 36 mins ago
José Carlos Santos
146k22116216
146k22116216
add a comment |
add a comment |
up vote
7
down vote
No that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.
Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
answered 36 mins ago
Yanko
5,432722
add a comment |
up vote
7
down vote
No that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.
Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
answered 36 mins ago
Yanko
5,432722
add a comment |
up vote
7
down vote
up vote
7
down vote
No that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.
Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
answered 36 mins ago
Yanko
5,432722
No that's wrong. The fact that a set is closed doesn't mean it is not open!
In fact $K$ is also open because it equals to $(sqrt{2},sqrt{3})cap mathbb{Q}$.
Side note: The space $mathbb{Q}$ with the topology induced by $mathbb{R}$ is "totally disconnected" this means that it has "many" sets which are both closed and open.
answered 36 mins ago
Yanko
5,432722
answered 36 mins ago
Yanko
5,432722
answered 36 mins ago
Yanko
5,432722
answered 36 mins ago
Yanko
5,432722
5,432722
add a comment |
add a comment |
up vote
1
down vote
A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.
Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.
answered 36 mins ago
Arthur
110k7104186
add a comment |
up vote
1
down vote
A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.
Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.
answered 36 mins ago
Arthur
110k7104186
add a comment |
up vote
1
down vote
up vote
1
down vote
A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.
Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.
answered 36 mins ago
Arthur
110k7104186
A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $[sqrt2, sqrt3]cap Bbb Q$ is closed in $Bbb Q$, that doen't mean it isn't open.
Look at the definition of open in the subspace topology, and se whether $[sqrt2, sqrt3]cap Bbb Q$ is such a set or not.
answered 36 mins ago
Arthur
110k7104186
answered 36 mins ago
Arthur
110k7104186
answered 36 mins ago
Arthur
110k7104186
answered 36 mins ago
Arthur
110k7104186
110k7104186
add a comment |
add a comment |
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