Why this topological space is compact?
$begingroup$
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
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
$endgroup$
|
show 2 more comments
$begingroup$
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
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
$endgroup$
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
|
show 2 more comments
$begingroup$
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
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
$endgroup$
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
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
asked Jan 21 at 17:24
deeppinkwaterdeeppinkwater
658
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
|
show 2 more comments
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
|
show 2 more comments
1 Answer
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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.
answered Jan 21 at 17:29
Paul KPaul K
2,765416
$endgroup$
$begingroup$
I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
$endgroup$
– idm
Jan 21 at 17:43
add a comment |
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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.
answered Jan 21 at 17:29
Paul KPaul K
2,765416
$endgroup$
$begingroup$
I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
$endgroup$
– idm
Jan 21 at 17:43
add a comment |
$begingroup$
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.
answered Jan 21 at 17:29
Paul KPaul K
2,765416
$endgroup$
$begingroup$
I thought that the topological space was $mathbb R$. Sorry. Good job, and nice answer (+1)
$endgroup$
– idm
Jan 21 at 17:43
add a comment |
$begingroup$
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.
answered Jan 21 at 17:29
Paul KPaul K
2,765416
$endgroup$
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.
answered Jan 21 at 17:29
Paul KPaul K
2,765416
edited Jan 21 at 17:34
answered Jan 21 at 17:29
Paul KPaul K
2,765416
answered Jan 21 at 17:29
Paul KPaul K
2,765416
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
add a comment |
$begingroup$
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
add a comment |
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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