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 yesterday
José Carlos Santos
140k18111203
asked yesterday
Messi fifa
45811
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 yesterday
José Carlos Santos
140k18111203
asked yesterday
Messi fifa
45811
1
It is not. You must think more carefully about compactness.
– John Douma
yesterday
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 yesterday
José Carlos Santos
140k18111203
asked yesterday
Messi fifa
45811
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 yesterday
José Carlos Santos
140k18111203
asked yesterday
Messi fifa
45811
edited yesterday
José Carlos Santos
140k18111203
asked yesterday
Messi fifa
45811
edited yesterday
José Carlos Santos
140k18111203
edited yesterday
José Carlos Santos
140k18111203
edited yesterday
José Carlos Santos
140k18111203
140k18111203
asked yesterday
Messi fifa
45811
asked yesterday
Messi fifa
45811
asked yesterday
Messi fifa
45811
45811
1
It is not. You must think more carefully about compactness.
– John Douma
yesterday
add a comment |
1
It is not. You must think more carefully about compactness.
– John Douma
yesterday
1
1
It is not. You must think more carefully about compactness.
– John Douma
yesterday
It is not. You must think more carefully about compactness.
– John Douma
yesterday
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 yesterday
José Carlos Santos
140k18111203
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
|
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 yesterday
José Carlos Santos
140k18111203
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
|
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 yesterday
José Carlos Santos
140k18111203
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
|
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 yesterday
José Carlos Santos
140k18111203
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 yesterday
José Carlos Santos
140k18111203
edited yesterday
answered yesterday
José Carlos Santos
140k18111203
answered yesterday
José Carlos Santos
140k18111203
answered yesterday
José Carlos Santos
140k18111203
140k18111203
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
|
show 3 more comments
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
It is closed in $mathbb Q$ because its complement is open.
– John Douma
yesterday
1
1
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
Right, 2 and 3 are correct.
– John Douma
yesterday
1
1
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
The set $A$ is open on $mathbb Q$.
– José Carlos Santos
yesterday
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
yesterday
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
yesterday
|
show 3 more comments
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006970%2flet-a-be-the-set-of-all-rational-p-such-that-2-p2-3-then-a-is-choos%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
It is not. You must think more carefully about compactness.
– John Douma
yesterday