Is my result and proof correct? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Show that $Z(p):pin mathscr P$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.Elementary topology question about bases and topologiesvery elementary question about bases on the real lineWhy does a topology contains its basis?Base for the Topology Generated by a Family of Semi-norms?Proof attempt for collection of all open intervals being a basis of $Bbb R$ with the standard topologyShow that $bigcup_n=1^infty A_n= B_1 backslash bigcap_n=1^infty B_n$Show basis for a topologyCompleting the converse of the theorem.Is my definition correct? How do I prove the finite union of elements of $mathscr U$ lie in $mathscr U$?Show that $Z(p):pin mathscr P$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.
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Is my result and proof correct?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Show that $Z(p):pin mathscr P$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.Elementary topology question about bases and topologiesvery elementary question about bases on the real lineWhy does a topology contains its basis?Base for the Topology Generated by a Family of Semi-norms?Proof attempt for collection of all open intervals being a basis of $Bbb R$ with the standard topologyShow that $bigcup_n=1^infty A_n= B_1 backslash bigcap_n=1^infty B_n$Show basis for a topologyCompleting the converse of the theorem.Is my definition correct? How do I prove the finite union of elements of $mathscr U$ lie in $mathscr U$?Show that $Z(p):pin mathscr P$ is a basis for the closed sets of some topology (Called the Zariski topology) on $mathbb R^n$.
$begingroup$
Let $(X,mathscr T)$ be a topological space. If $mathscr B=B_alpha:alpha in Lambda$ is a collection of closed sets that forms a base for the closed subsets of $X$. Will $mathscr D=B^c:Bin mathscr B$ form a base for $mathscr T$?
My Attempt.
Claim:-$mathscr D=B^c:Bin mathscr B$ forms a basis for $mathscr T$
(1) $bigcup_Bin mathscr BB^c=(bigcap_alpha in LambdaB_alpha)^c=emptyset^c=X$
(2)$B_1^c,B_2^c in mathscr D$ and $xin B_1^c cap B_2^cimplies xin (B_1 cup B_2)^c $. Then there is $B_3in mathscr B: xnotin B_3supseteq B_1 cup B_2$. Taking the complement on Both sides. We get $xin B_3^csubseteq B_1^c cap B_2^c$. Hence, $mathscr D$ forms a basis for $mathscr T$.
Is my result and proof correct?
general-topology
edited Apr 2 at 4:44
Henno Brandsma
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
$endgroup$
add a comment |
$begingroup$
Let $(X,mathscr T)$ be a topological space. If $mathscr B=B_alpha:alpha in Lambda$ is a collection of closed sets that forms a base for the closed subsets of $X$. Will $mathscr D=B^c:Bin mathscr B$ form a base for $mathscr T$?
My Attempt.
Claim:-$mathscr D=B^c:Bin mathscr B$ forms a basis for $mathscr T$
(1) $bigcup_Bin mathscr BB^c=(bigcap_alpha in LambdaB_alpha)^c=emptyset^c=X$
(2)$B_1^c,B_2^c in mathscr D$ and $xin B_1^c cap B_2^cimplies xin (B_1 cup B_2)^c $. Then there is $B_3in mathscr B: xnotin B_3supseteq B_1 cup B_2$. Taking the complement on Both sides. We get $xin B_3^csubseteq B_1^c cap B_2^c$. Hence, $mathscr D$ forms a basis for $mathscr T$.
Is my result and proof correct?
general-topology
edited Apr 2 at 4:44
Henno Brandsma
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
$endgroup$
$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28
add a comment |
$begingroup$
Let $(X,mathscr T)$ be a topological space. If $mathscr B=B_alpha:alpha in Lambda$ is a collection of closed sets that forms a base for the closed subsets of $X$. Will $mathscr D=B^c:Bin mathscr B$ form a base for $mathscr T$?
My Attempt.
Claim:-$mathscr D=B^c:Bin mathscr B$ forms a basis for $mathscr T$
(1) $bigcup_Bin mathscr BB^c=(bigcap_alpha in LambdaB_alpha)^c=emptyset^c=X$
(2)$B_1^c,B_2^c in mathscr D$ and $xin B_1^c cap B_2^cimplies xin (B_1 cup B_2)^c $. Then there is $B_3in mathscr B: xnotin B_3supseteq B_1 cup B_2$. Taking the complement on Both sides. We get $xin B_3^csubseteq B_1^c cap B_2^c$. Hence, $mathscr D$ forms a basis for $mathscr T$.
Is my result and proof correct?
general-topology
edited Apr 2 at 4:44
Henno Brandsma
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
$endgroup$
Let $(X,mathscr T)$ be a topological space. If $mathscr B=B_alpha:alpha in Lambda$ is a collection of closed sets that forms a base for the closed subsets of $X$. Will $mathscr D=B^c:Bin mathscr B$ form a base for $mathscr T$?
My Attempt.
Claim:-$mathscr D=B^c:Bin mathscr B$ forms a basis for $mathscr T$
(1) $bigcup_Bin mathscr BB^c=(bigcap_alpha in LambdaB_alpha)^c=emptyset^c=X$
(2)$B_1^c,B_2^c in mathscr D$ and $xin B_1^c cap B_2^cimplies xin (B_1 cup B_2)^c $. Then there is $B_3in mathscr B: xnotin B_3supseteq B_1 cup B_2$. Taking the complement on Both sides. We get $xin B_3^csubseteq B_1^c cap B_2^c$. Hence, $mathscr D$ forms a basis for $mathscr T$.
Is my result and proof correct?
general-topology
general-topology
edited Apr 2 at 4:44
Henno Brandsma
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
edited Apr 2 at 4:44
Henno Brandsma
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
edited Apr 2 at 4:44
Henno Brandsma
117k350128
edited Apr 2 at 4:44
Henno Brandsma
117k350128
edited Apr 2 at 4:44
Henno Brandsma
117k350128
117k350128
asked Apr 2 at 1:09
Math geekMath geek
69111
asked Apr 2 at 1:09
Math geekMath geek
69111
asked Apr 2 at 1:09
Math geekMath geek
69111
69111
$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28
add a comment |
$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28
$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The topology is given, so note that in a base for the closed sets all $mathcalT$-closed sets are intersection of subfamilies of $mathcalB$, so by de Morgan all $mathcalT$-open sets are unions of their complements, and $B^complement: B in mathcalB$ thus forms a base for $mathcalT$.
You are checking the general precondition for forming a possible base in general, not for this topology specifically.
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
The topology is given, so note that in a base for the closed sets all $mathcalT$-closed sets are intersection of subfamilies of $mathcalB$, so by de Morgan all $mathcalT$-open sets are unions of their complements, and $B^complement: B in mathcalB$ thus forms a base for $mathcalT$.
You are checking the general precondition for forming a possible base in general, not for this topology specifically.
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
The topology is given, so note that in a base for the closed sets all $mathcalT$-closed sets are intersection of subfamilies of $mathcalB$, so by de Morgan all $mathcalT$-open sets are unions of their complements, and $B^complement: B in mathcalB$ thus forms a base for $mathcalT$.
You are checking the general precondition for forming a possible base in general, not for this topology specifically.
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
The topology is given, so note that in a base for the closed sets all $mathcalT$-closed sets are intersection of subfamilies of $mathcalB$, so by de Morgan all $mathcalT$-open sets are unions of their complements, and $B^complement: B in mathcalB$ thus forms a base for $mathcalT$.
You are checking the general precondition for forming a possible base in general, not for this topology specifically.
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
The topology is given, so note that in a base for the closed sets all $mathcalT$-closed sets are intersection of subfamilies of $mathcalB$, so by de Morgan all $mathcalT$-open sets are unions of their complements, and $B^complement: B in mathcalB$ thus forms a base for $mathcalT$.
You are checking the general precondition for forming a possible base in general, not for this topology specifically.
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
edited Apr 2 at 5:50
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
answered Apr 2 at 4:48
Henno BrandsmaHenno Brandsma
117k350128
117k350128
add a comment |
add a comment |
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$begingroup$
No. You omitted a requirement of a topological space,
$endgroup$
– William Elliot
Apr 2 at 2:13
$begingroup$
can you please explain? what did I miss?
$endgroup$
– Math geek
Apr 2 at 5:25
$begingroup$
Please for the next time, add what the question is about on the title, so that people can know beforehand whether they will be able to help or not.
$endgroup$
– David
Apr 2 at 7:28