Why this topological space is compact?












0












$begingroup$


Let $X = [0, + infty[ subset mathbb{R}$ and $tau$ be a topology on $X$ which is composed by $X$, $emptyset$ and every set of the form $]a ; + infty[$, where $a in mathbb{R}$.



Can somebody explain to me why $(X, tau)$ is a compact topological space ?



If I take the family of set $(U_{i})_{i in mathbb{N}}$ such that for every $i in mathbb{N}$, $U_{i} = ]frac{1}{n}, + infty[$, then we have that $bigcup_{i in mathbb{N}} U_{i} = X$ (so $(U_{i})_{i in mathbb{N}}$ is an open cover of $X$), but $(U_{i})_{i in mathbb{N}}$ has no finit subcover of $X$ right (or I'm probably missing something) ?



Thank you for your help.






general-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The union of your $U_i$ does not contain $0$.
    $endgroup$
    – Paul K
    Jan 21 at 17:27










  • $begingroup$
    The only open set that contains $0$ is $X$.
    $endgroup$
    – Henno Brandsma
    Jan 21 at 17:30










  • $begingroup$
    So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:30








  • 1




    $begingroup$
    Indeed, $X’$ is not compact anymore.
    $endgroup$
    – Mindlack
    Jan 21 at 17:34












  • $begingroup$
    Thank you for your help !
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:35
















0












$begingroup$


Let $X = [0, + infty[ subset mathbb{R}$ and $tau$ be a topology on $X$ which is composed by $X$, $emptyset$ and every set of the form $]a ; + infty[$, where $a in mathbb{R}$.



Can somebody explain to me why $(X, tau)$ is a compact topological space ?



If I take the family of set $(U_{i})_{i in mathbb{N}}$ such that for every $i in mathbb{N}$, $U_{i} = ]frac{1}{n}, + infty[$, then we have that $bigcup_{i in mathbb{N}} U_{i} = X$ (so $(U_{i})_{i in mathbb{N}}$ is an open cover of $X$), but $(U_{i})_{i in mathbb{N}}$ has no finit subcover of $X$ right (or I'm probably missing something) ?



Thank you for your help.






general-topology







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The union of your $U_i$ does not contain $0$.
    $endgroup$
    – Paul K
    Jan 21 at 17:27










  • $begingroup$
    The only open set that contains $0$ is $X$.
    $endgroup$
    – Henno Brandsma
    Jan 21 at 17:30










  • $begingroup$
    So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:30








  • 1




    $begingroup$
    Indeed, $X’$ is not compact anymore.
    $endgroup$
    – Mindlack
    Jan 21 at 17:34












  • $begingroup$
    Thank you for your help !
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:35














0












0








0





$begingroup$


Let $X = [0, + infty[ subset mathbb{R}$ and $tau$ be a topology on $X$ which is composed by $X$, $emptyset$ and every set of the form $]a ; + infty[$, where $a in mathbb{R}$.



Can somebody explain to me why $(X, tau)$ is a compact topological space ?



If I take the family of set $(U_{i})_{i in mathbb{N}}$ such that for every $i in mathbb{N}$, $U_{i} = ]frac{1}{n}, + infty[$, then we have that $bigcup_{i in mathbb{N}} U_{i} = X$ (so $(U_{i})_{i in mathbb{N}}$ is an open cover of $X$), but $(U_{i})_{i in mathbb{N}}$ has no finit subcover of $X$ right (or I'm probably missing something) ?



Thank you for your help.






general-topology







share|cite|improve this question









$endgroup$




Let $X = [0, + infty[ subset mathbb{R}$ and $tau$ be a topology on $X$ which is composed by $X$, $emptyset$ and every set of the form $]a ; + infty[$, where $a in mathbb{R}$.



Can somebody explain to me why $(X, tau)$ is a compact topological space ?



If I take the family of set $(U_{i})_{i in mathbb{N}}$ such that for every $i in mathbb{N}$, $U_{i} = ]frac{1}{n}, + infty[$, then we have that $bigcup_{i in mathbb{N}} U_{i} = X$ (so $(U_{i})_{i in mathbb{N}}$ is an open cover of $X$), but $(U_{i})_{i in mathbb{N}}$ has no finit subcover of $X$ right (or I'm probably missing something) ?



Thank you for your help.






general-topology




general-topology






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 21 at 17:24









deeppinkwaterdeeppinkwater

658




658








  • 1




    $begingroup$
    The union of your $U_i$ does not contain $0$.
    $endgroup$
    – Paul K
    Jan 21 at 17:27










  • $begingroup$
    The only open set that contains $0$ is $X$.
    $endgroup$
    – Henno Brandsma
    Jan 21 at 17:30










  • $begingroup$
    So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:30








  • 1




    $begingroup$
    Indeed, $X’$ is not compact anymore.
    $endgroup$
    – Mindlack
    Jan 21 at 17:34












  • $begingroup$
    Thank you for your help !
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:35














  • 1




    $begingroup$
    The union of your $U_i$ does not contain $0$.
    $endgroup$
    – Paul K
    Jan 21 at 17:27










  • $begingroup$
    The only open set that contains $0$ is $X$.
    $endgroup$
    – Henno Brandsma
    Jan 21 at 17:30










  • $begingroup$
    So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:30








  • 1




    $begingroup$
    Indeed, $X’$ is not compact anymore.
    $endgroup$
    – Mindlack
    Jan 21 at 17:34












  • $begingroup$
    Thank you for your help !
    $endgroup$
    – deeppinkwater
    Jan 21 at 17:35








1




1




$begingroup$
The union of your $U_i$ does not contain $0$.
$endgroup$
– Paul K
Jan 21 at 17:27




$begingroup$
The union of your $U_i$ does not contain $0$.
$endgroup$
– Paul K
Jan 21 at 17:27












$begingroup$
The only open set that contains $0$ is $X$.
$endgroup$
– Henno Brandsma
Jan 21 at 17:30




$begingroup$
The only open set that contains $0$ is $X$.
$endgroup$
– Henno Brandsma
Jan 21 at 17:30












$begingroup$
So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
$endgroup$
– deeppinkwater
Jan 21 at 17:30






$begingroup$
So if I replace $X$ by $X' = ]0, + infty[$ (keeping the 'same' topology), then $X'$ is not compact (taking my example) ?
$endgroup$
– deeppinkwater
Jan 21 at 17:30






1




1




$begingroup$
Indeed, $X’$ is not compact anymore.
$endgroup$
– Mindlack
Jan 21 at 17:34






$begingroup$
Indeed, $X’$ is not compact anymore.
$endgroup$
– Mindlack
Jan 21 at 17:34














$begingroup$
Thank you for your help !
$endgroup$
– deeppinkwater
Jan 21 at 17:35




$begingroup$
Thank you for your help !
$endgroup$
– deeppinkwater
Jan 21 at 17:35










1 Answer
1






active

oldest

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2












$begingroup$

The union of your sets $U_i$ is only the interval $(0, infty)$. To show that $X$ is compact, take some open cover $U_i$ with $bigcup U_i = X$. Since $0 in X$, there is some $U_j$ with $0 in U_j$. But then $U_j = X = [0, infty)$ and thus ${U_j}$ is a finite subcover.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
    $endgroup$
    – idm
    Jan 21 at 17:43













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1 Answer
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oldest

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active

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active

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$begingroup$

The union of your sets $U_i$ is only the interval $(0, infty)$. To show that $X$ is compact, take some open cover $U_i$ with $bigcup U_i = X$. Since $0 in X$, there is some $U_j$ with $0 in U_j$. But then $U_j = X = [0, infty)$ and thus ${U_j}$ is a finite subcover.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
    $endgroup$
    – idm
    Jan 21 at 17:43


















2












$begingroup$

The union of your sets $U_i$ is only the interval $(0, infty)$. To show that $X$ is compact, take some open cover $U_i$ with $bigcup U_i = X$. Since $0 in X$, there is some $U_j$ with $0 in U_j$. But then $U_j = X = [0, infty)$ and thus ${U_j}$ is a finite subcover.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
    $endgroup$
    – idm
    Jan 21 at 17:43
















2












2








2





$begingroup$

The union of your sets $U_i$ is only the interval $(0, infty)$. To show that $X$ is compact, take some open cover $U_i$ with $bigcup U_i = X$. Since $0 in X$, there is some $U_j$ with $0 in U_j$. But then $U_j = X = [0, infty)$ and thus ${U_j}$ is a finite subcover.






share|cite|improve this answer











$endgroup$



The union of your sets $U_i$ is only the interval $(0, infty)$. To show that $X$ is compact, take some open cover $U_i$ with $bigcup U_i = X$. Since $0 in X$, there is some $U_j$ with $0 in U_j$. But then $U_j = X = [0, infty)$ and thus ${U_j}$ is a finite subcover.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 21 at 17:34

























answered Jan 21 at 17:29









Paul KPaul K

2,765416




2,765416












  • $begingroup$
    I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
    $endgroup$
    – idm
    Jan 21 at 17:43




















  • $begingroup$
    I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
    $endgroup$
    – idm
    Jan 21 at 17:43


















$begingroup$
I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
$endgroup$
– idm
Jan 21 at 17:43






$begingroup$
I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
$endgroup$
– idm
Jan 21 at 17:43




















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