Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is choose the correct option
up vote
0
down vote
favorite
Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is
choose the correct option
$1)$ compact in $mathbb{Q}$
$2)$closed and bounded in $mathbb{Q}$
$3)$Not compact in $mathbb{Q}$
$4)$ closed and unbounded in $mathbb{Q}$
i thinks option $1)$ and $2)$ option will be correct by Heine Boral theorem
Is its True ?
real-analysis metric-spaces compactness
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
add a comment |
up vote
0
down vote
favorite
Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is
choose the correct option
$1)$ compact in $mathbb{Q}$
$2)$closed and bounded in $mathbb{Q}$
$3)$Not compact in $mathbb{Q}$
$4)$ closed and unbounded in $mathbb{Q}$
i thinks option $1)$ and $2)$ option will be correct by Heine Boral theorem
Is its True ?
real-analysis metric-spaces compactness
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
1
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is
choose the correct option
$1)$ compact in $mathbb{Q}$
$2)$closed and bounded in $mathbb{Q}$
$3)$Not compact in $mathbb{Q}$
$4)$ closed and unbounded in $mathbb{Q}$
i thinks option $1)$ and $2)$ option will be correct by Heine Boral theorem
Is its True ?
real-analysis metric-spaces compactness
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is
choose the correct option
$1)$ compact in $mathbb{Q}$
$2)$closed and bounded in $mathbb{Q}$
$3)$Not compact in $mathbb{Q}$
$4)$ closed and unbounded in $mathbb{Q}$
i thinks option $1)$ and $2)$ option will be correct by Heine Boral theorem
Is its True ?
real-analysis metric-spaces compactness
real-analysis metric-spaces compactness
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
edited Nov 20 at 22:25
José Carlos Santos
144k20112212
144k20112212
asked Nov 20 at 22:10
Messi fifa
50111
asked Nov 20 at 22:10
Messi fifa
50111
asked Nov 20 at 22:10
Messi fifa
50111
50111
1
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16
add a comment |
1
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16
1
1
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The set $A$ is not closed in $mathbb R$. For instance, $sqrt2$ belongs to $overline A$, but not to $A$. So, $A$ is not compact and 1) doesn't hold. Besides, $A$ is clearly bounded and also closed in $mathbb Q$, so 2) holds and 4) also hold. The correct choices are 2) and 3). If you want to show directly that $A$ is not compact (in $mathbb Q$ or in $mathbb R$, it doesn't matter), use the fact that it is not a closed set.
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
s/also/doesn't/(typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?
– egreg
Nov 20 at 22:53
|
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The set $A$ is not closed in $mathbb R$. For instance, $sqrt2$ belongs to $overline A$, but not to $A$. So, $A$ is not compact and 1) doesn't hold. Besides, $A$ is clearly bounded and also closed in $mathbb Q$, so 2) holds and 4) also hold. The correct choices are 2) and 3). If you want to show directly that $A$ is not compact (in $mathbb Q$ or in $mathbb R$, it doesn't matter), use the fact that it is not a closed set.
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
s/also/doesn't/(typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?
– egreg
Nov 20 at 22:53
|
show 3 more comments
up vote
1
down vote
accepted
The set $A$ is not closed in $mathbb R$. For instance, $sqrt2$ belongs to $overline A$, but not to $A$. So, $A$ is not compact and 1) doesn't hold. Besides, $A$ is clearly bounded and also closed in $mathbb Q$, so 2) holds and 4) also hold. The correct choices are 2) and 3). If you want to show directly that $A$ is not compact (in $mathbb Q$ or in $mathbb R$, it doesn't matter), use the fact that it is not a closed set.
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
s/also/doesn't/(typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?
– egreg
Nov 20 at 22:53
|
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The set $A$ is not closed in $mathbb R$. For instance, $sqrt2$ belongs to $overline A$, but not to $A$. So, $A$ is not compact and 1) doesn't hold. Besides, $A$ is clearly bounded and also closed in $mathbb Q$, so 2) holds and 4) also hold. The correct choices are 2) and 3). If you want to show directly that $A$ is not compact (in $mathbb Q$ or in $mathbb R$, it doesn't matter), use the fact that it is not a closed set.
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
The set $A$ is not closed in $mathbb R$. For instance, $sqrt2$ belongs to $overline A$, but not to $A$. So, $A$ is not compact and 1) doesn't hold. Besides, $A$ is clearly bounded and also closed in $mathbb Q$, so 2) holds and 4) also hold. The correct choices are 2) and 3). If you want to show directly that $A$ is not compact (in $mathbb Q$ or in $mathbb R$, it doesn't matter), use the fact that it is not a closed set.
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
edited Nov 20 at 22:33
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
answered Nov 20 at 22:25
José Carlos Santos
144k20112212
144k20112212
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
s/also/doesn't/(typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?
– egreg
Nov 20 at 22:53
|
show 3 more comments
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
s/also/doesn't/(typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?
– egreg
Nov 20 at 22:53
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 at 22:29
1
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 at 22:30
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
Right, 2 and 3 are correct.
– John Douma
Nov 20 at 22:31
1
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 at 22:48
1
1
s/also/doesn't/ (typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?– egreg
Nov 20 at 22:53
s/also/doesn't/ (typo). Be careful: $A$ is closed in $mathbb{Q}$: can you point to a rational which is an accumulation point and doesn't belong to $A$?– egreg
Nov 20 at 22:53
|
show 3 more comments
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1
It is not. You must think more carefully about compactness.
– John Douma
Nov 20 at 22:16