About isomorphic of a boolean algebra to bolean algebra
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Let $mathcal A$ be a boolean algebra. It`s non-empty subset $mathcal F$ is called a filter if $∅ notin mathcal F$, for all $A ∈ mathcal A$, $B ∈ mathcal F$ from $B ⊂ A$ follows $A ∈ mathcal F$, and for all $A, B ∈ mathcal F$ follows
$A∩B ∈ mathcal F$. A filter is said to be maximal if for all $A ∈ mathcal A$ or $A ∈ mathcal F$, or $ overline A ∈ mathcal F$. How to show that
on the set of all maximal filters one can construct a Boolean algebra isomorphic to the original Boolean algebra $mathcal A$?
abstract-algebra boolean-algebra filters
asked Dec 11 '18 at 5:15
LisaLisa
275
$endgroup$
add a comment |
$begingroup$
Let $mathcal A$ be a boolean algebra. It`s non-empty subset $mathcal F$ is called a filter if $∅ notin mathcal F$, for all $A ∈ mathcal A$, $B ∈ mathcal F$ from $B ⊂ A$ follows $A ∈ mathcal F$, and for all $A, B ∈ mathcal F$ follows
$A∩B ∈ mathcal F$. A filter is said to be maximal if for all $A ∈ mathcal A$ or $A ∈ mathcal F$, or $ overline A ∈ mathcal F$. How to show that
on the set of all maximal filters one can construct a Boolean algebra isomorphic to the original Boolean algebra $mathcal A$?
abstract-algebra boolean-algebra filters
asked Dec 11 '18 at 5:15
LisaLisa
275
$endgroup$
1
$begingroup$
read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34
add a comment |
$begingroup$
Let $mathcal A$ be a boolean algebra. It`s non-empty subset $mathcal F$ is called a filter if $∅ notin mathcal F$, for all $A ∈ mathcal A$, $B ∈ mathcal F$ from $B ⊂ A$ follows $A ∈ mathcal F$, and for all $A, B ∈ mathcal F$ follows
$A∩B ∈ mathcal F$. A filter is said to be maximal if for all $A ∈ mathcal A$ or $A ∈ mathcal F$, or $ overline A ∈ mathcal F$. How to show that
on the set of all maximal filters one can construct a Boolean algebra isomorphic to the original Boolean algebra $mathcal A$?
abstract-algebra boolean-algebra filters
asked Dec 11 '18 at 5:15
LisaLisa
275
$endgroup$
Let $mathcal A$ be a boolean algebra. It`s non-empty subset $mathcal F$ is called a filter if $∅ notin mathcal F$, for all $A ∈ mathcal A$, $B ∈ mathcal F$ from $B ⊂ A$ follows $A ∈ mathcal F$, and for all $A, B ∈ mathcal F$ follows
$A∩B ∈ mathcal F$. A filter is said to be maximal if for all $A ∈ mathcal A$ or $A ∈ mathcal F$, or $ overline A ∈ mathcal F$. How to show that
on the set of all maximal filters one can construct a Boolean algebra isomorphic to the original Boolean algebra $mathcal A$?
abstract-algebra boolean-algebra filters
abstract-algebra boolean-algebra filters
asked Dec 11 '18 at 5:15
LisaLisa
275
asked Dec 11 '18 at 5:15
LisaLisa
275
asked Dec 11 '18 at 5:15
LisaLisa
275
asked Dec 11 '18 at 5:15
LisaLisa
275
asked Dec 11 '18 at 5:15
LisaLisa
275
275
1
$begingroup$
read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34
add a comment |
1
$begingroup$
read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34
1
1
$begingroup$
read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34
$begingroup$
read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34
add a comment |
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read up on Stone duality. e.g. chapter 2 here
$endgroup$
– Henno Brandsma
Dec 11 '18 at 5:34