Locally compactness
I want to show that subspace of $mathbb R$ is locally compact.
Let $W$ be subspace of $mathbb R$. Let $xin W$ and $yin mathbb R setminus W$. Since $mathbb R$ is Hausdorff space, there exists disjoint open sets $U$ and $V$ such that $xin U$ and $yin V$ for $xin W$, $yin mathbb R setminus W$. Then, $Usubseteq X setminus V$ and $Xsetminus V$ is closed in $mathbb R$, so the closure of $U$ is contained in $X setminus V $. Since the closure of $U$ is bounded and closed in $mathbb R$, the closure is compact. Hence, there exists open set containing $x$ such that compact subset is contained in the open set. That means $W$ is locally compact.
Please let me know whether it's right or not.
general-topology
asked Dec 3 '18 at 8:07
dlfjsemf
968
add a comment |
I want to show that subspace of $mathbb R$ is locally compact.
Let $W$ be subspace of $mathbb R$. Let $xin W$ and $yin mathbb R setminus W$. Since $mathbb R$ is Hausdorff space, there exists disjoint open sets $U$ and $V$ such that $xin U$ and $yin V$ for $xin W$, $yin mathbb R setminus W$. Then, $Usubseteq X setminus V$ and $Xsetminus V$ is closed in $mathbb R$, so the closure of $U$ is contained in $X setminus V $. Since the closure of $U$ is bounded and closed in $mathbb R$, the closure is compact. Hence, there exists open set containing $x$ such that compact subset is contained in the open set. That means $W$ is locally compact.
Please let me know whether it's right or not.
general-topology
asked Dec 3 '18 at 8:07
dlfjsemf
968
You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13
add a comment |
I want to show that subspace of $mathbb R$ is locally compact.
Let $W$ be subspace of $mathbb R$. Let $xin W$ and $yin mathbb R setminus W$. Since $mathbb R$ is Hausdorff space, there exists disjoint open sets $U$ and $V$ such that $xin U$ and $yin V$ for $xin W$, $yin mathbb R setminus W$. Then, $Usubseteq X setminus V$ and $Xsetminus V$ is closed in $mathbb R$, so the closure of $U$ is contained in $X setminus V $. Since the closure of $U$ is bounded and closed in $mathbb R$, the closure is compact. Hence, there exists open set containing $x$ such that compact subset is contained in the open set. That means $W$ is locally compact.
Please let me know whether it's right or not.
general-topology
asked Dec 3 '18 at 8:07
dlfjsemf
968
I want to show that subspace of $mathbb R$ is locally compact.
Let $W$ be subspace of $mathbb R$. Let $xin W$ and $yin mathbb R setminus W$. Since $mathbb R$ is Hausdorff space, there exists disjoint open sets $U$ and $V$ such that $xin U$ and $yin V$ for $xin W$, $yin mathbb R setminus W$. Then, $Usubseteq X setminus V$ and $Xsetminus V$ is closed in $mathbb R$, so the closure of $U$ is contained in $X setminus V $. Since the closure of $U$ is bounded and closed in $mathbb R$, the closure is compact. Hence, there exists open set containing $x$ such that compact subset is contained in the open set. That means $W$ is locally compact.
Please let me know whether it's right or not.
general-topology
general-topology
asked Dec 3 '18 at 8:07
dlfjsemf
968
asked Dec 3 '18 at 8:07
dlfjsemf
968
asked Dec 3 '18 at 8:07
dlfjsemf
968
asked Dec 3 '18 at 8:07
dlfjsemf
968
asked Dec 3 '18 at 8:07
dlfjsemf
968
968
You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13
add a comment |
You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13
You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13
You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13
add a comment |
1 Answer
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It is well known (and not hard to prove) that $mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
add a comment |
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It is well known (and not hard to prove) that $mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
add a comment |
It is well known (and not hard to prove) that $mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
add a comment |
It is well known (and not hard to prove) that $mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
It is well known (and not hard to prove) that $mathbb Q$ is not locally compact. See my comment about the mistakes in your argument.
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
answered Dec 3 '18 at 8:16
Kavi Rama Murthy
50.4k31854
50.4k31854
add a comment |
add a comment |
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You have to produce a neighborhood of $x$ in the subspace topology of $W$ whose closure in $W$ is compact. Your argument does not involve the relative topology on $W$ so it is not correct.
– Kavi Rama Murthy
Dec 3 '18 at 8:13