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Is my result and proof correct?

1

$begingroup$

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

share|cite|improve this question

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 | 

1

$begingroup$

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

share|cite|improve this question

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

share|cite|improve this question

edited Apr 2 at 4:44

Henno Brandsma

117k350128

asked Apr 2 at 1:09

Math geekMath geek

69111

$endgroup$

share|cite|improve this question

edited Apr 2 at 4:44

Henno Brandsma

117k350128

asked Apr 2 at 1:09

Math geekMath geek

69111

share|cite|improve this question

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

asked Apr 2 at 1:09

Math geekMath geek

69111

asked Apr 2 at 1:09

Math geekMath geek

69111

1 Answer
1

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:48

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

add a comment | 

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2

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:48

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

edited Apr 2 at 5:50

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.

share|cite|improve this answer

edited Apr 2 at 5:50

answered Apr 2 at 4:48

Henno BrandsmaHenno Brandsma

117k350128

$endgroup$

share|cite|improve this answer

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