Let $A$ be the set of all rational $p$ such that $2 < p^2 < 3$.Then $A$ is choose the correct option












0














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







share|cite|improve this question




















  • 1




    It is not. You must think more carefully about compactness.
    – John Douma
    Nov 20 '18 at 22:16
















0














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







share|cite|improve this question




















  • 1




    It is not. You must think more carefully about compactness.
    – John Douma
    Nov 20 '18 at 22:16














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 22:25









José Carlos Santos

150k22121221




150k22121221










asked Nov 20 '18 at 22:10









Messi fifa

51611




51611








  • 1




    It is not. You must think more carefully about compactness.
    – John Douma
    Nov 20 '18 at 22:16














  • 1




    It is not. You must think more carefully about compactness.
    – John Douma
    Nov 20 '18 at 22:16








1




1




It is not. You must think more carefully about compactness.
– John Douma
Nov 20 '18 at 22:16




It is not. You must think more carefully about compactness.
– John Douma
Nov 20 '18 at 22:16










1 Answer
1






active

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1














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.






share|cite|improve this answer























  • It is closed in $mathbb Q$ because its complement is open.
    – John Douma
    Nov 20 '18 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 '18 at 22:30










  • Right, 2 and 3 are correct.
    – John Douma
    Nov 20 '18 at 22:31






  • 1




    The set $A$ is open on $mathbb Q$.
    – José Carlos Santos
    Nov 20 '18 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 '18 at 22:53













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1 Answer
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1 Answer
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active

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active

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1














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.






share|cite|improve this answer























  • It is closed in $mathbb Q$ because its complement is open.
    – John Douma
    Nov 20 '18 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 '18 at 22:30










  • Right, 2 and 3 are correct.
    – John Douma
    Nov 20 '18 at 22:31






  • 1




    The set $A$ is open on $mathbb Q$.
    – José Carlos Santos
    Nov 20 '18 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 '18 at 22:53


















1














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.






share|cite|improve this answer























  • It is closed in $mathbb Q$ because its complement is open.
    – John Douma
    Nov 20 '18 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 '18 at 22:30










  • Right, 2 and 3 are correct.
    – John Douma
    Nov 20 '18 at 22:31






  • 1




    The set $A$ is open on $mathbb Q$.
    – José Carlos Santos
    Nov 20 '18 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 '18 at 22:53
















1












1








1






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 20 '18 at 22:33

























answered Nov 20 '18 at 22:25









José Carlos Santos

150k22121221




150k22121221












  • It is closed in $mathbb Q$ because its complement is open.
    – John Douma
    Nov 20 '18 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 '18 at 22:30










  • Right, 2 and 3 are correct.
    – John Douma
    Nov 20 '18 at 22:31






  • 1




    The set $A$ is open on $mathbb Q$.
    – José Carlos Santos
    Nov 20 '18 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 '18 at 22:53




















  • It is closed in $mathbb Q$ because its complement is open.
    – John Douma
    Nov 20 '18 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 '18 at 22:30










  • Right, 2 and 3 are correct.
    – John Douma
    Nov 20 '18 at 22:31






  • 1




    The set $A$ is open on $mathbb Q$.
    – José Carlos Santos
    Nov 20 '18 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 '18 at 22:53


















It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 '18 at 22:29






It is closed in $mathbb Q$ because its complement is open.
– John Douma
Nov 20 '18 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 '18 at 22:30




But it is not a closed subset of $mathbb R$ and therefore it is not compact.
– José Carlos Santos
Nov 20 '18 at 22:30












Right, 2 and 3 are correct.
– John Douma
Nov 20 '18 at 22:31




Right, 2 and 3 are correct.
– John Douma
Nov 20 '18 at 22:31




1




1




The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 '18 at 22:48




The set $A$ is open on $mathbb Q$.
– José Carlos Santos
Nov 20 '18 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 '18 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 '18 at 22:53




















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