name of algebraic structure generalizing closure and interior
$begingroup$
Is there a name for an algebraic structure equipped with an involution and two additional unary functions that "act like" the interior and closure of a set?
I'm after something a little bit like an $i$-lattice / DeMorgan algebra but additionally with something that "feels like" $square$ and $lozenge$ from modal logic.
Suppose we take the sets of subsets of $mathbb{R}$ and consider the operations $cup(cdot, cdot);,; cap(cdot, cdot);,; cdot^complement;,; mathrm{int}(cdot);,; overline{(cdot)} $ , so union, intersection, complement, interior, and closure.
$cup$ and $cap$ distribute over each other and are duals with respect to $cdot^complement$ and $x^complement{^complement} = x$ .
Interior are closure are also dual and each is idempotent.
$$ mathrm{int}(x^complement)^complement = overline{x} tag{1} $$
$$ mathrm{int}(mathrm{int}(x)) = mathrm{int}(x) tag{2} $$
And the union of the closures is the closure of the (finite) union(s).
$$ overline{x} cup overline{y} = overline{x cup y} tag{3} $$
Here's an example of a finite structure satisfying the rules. The domain of our structure is ${text{-1}, 0, 1}$
$$ x cap y stackrel{df}{=} min(x, y) tag{4} $$
$$ x cup y stackrel{df}{=} max(x, y) tag{5} $$
$$ x ^complement stackrel{df}{=} -x tag{6} $$
$$ mathrm{int}(text{-1}) stackrel{df}{=} text{-1} \
mathrm{int}(0) stackrel{df}{=} 0 \
mathrm{int}(1) stackrel{df}{=} 0 tag{7} $$
$$ overline{(text{-1})} stackrel{df}{=} 0 \
overline{0} stackrel{df}{=} 0 \
overline{1} stackrel{df}{=} 1 tag{8} $$
Note that it isn't possible to swap the definitions in (7) and (8) because of (3) ... which guarantees that union and closure "move things in a similar direction".
abstract-algebra
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
$endgroup$
add a comment |
$begingroup$
Is there a name for an algebraic structure equipped with an involution and two additional unary functions that "act like" the interior and closure of a set?
I'm after something a little bit like an $i$-lattice / DeMorgan algebra but additionally with something that "feels like" $square$ and $lozenge$ from modal logic.
Suppose we take the sets of subsets of $mathbb{R}$ and consider the operations $cup(cdot, cdot);,; cap(cdot, cdot);,; cdot^complement;,; mathrm{int}(cdot);,; overline{(cdot)} $ , so union, intersection, complement, interior, and closure.
$cup$ and $cap$ distribute over each other and are duals with respect to $cdot^complement$ and $x^complement{^complement} = x$ .
Interior are closure are also dual and each is idempotent.
$$ mathrm{int}(x^complement)^complement = overline{x} tag{1} $$
$$ mathrm{int}(mathrm{int}(x)) = mathrm{int}(x) tag{2} $$
And the union of the closures is the closure of the (finite) union(s).
$$ overline{x} cup overline{y} = overline{x cup y} tag{3} $$
Here's an example of a finite structure satisfying the rules. The domain of our structure is ${text{-1}, 0, 1}$
$$ x cap y stackrel{df}{=} min(x, y) tag{4} $$
$$ x cup y stackrel{df}{=} max(x, y) tag{5} $$
$$ x ^complement stackrel{df}{=} -x tag{6} $$
$$ mathrm{int}(text{-1}) stackrel{df}{=} text{-1} \
mathrm{int}(0) stackrel{df}{=} 0 \
mathrm{int}(1) stackrel{df}{=} 0 tag{7} $$
$$ overline{(text{-1})} stackrel{df}{=} 0 \
overline{0} stackrel{df}{=} 0 \
overline{1} stackrel{df}{=} 1 tag{8} $$
Note that it isn't possible to swap the definitions in (7) and (8) because of (3) ... which guarantees that union and closure "move things in a similar direction".
abstract-algebra
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
$endgroup$
$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22
add a comment |
$begingroup$
Is there a name for an algebraic structure equipped with an involution and two additional unary functions that "act like" the interior and closure of a set?
I'm after something a little bit like an $i$-lattice / DeMorgan algebra but additionally with something that "feels like" $square$ and $lozenge$ from modal logic.
Suppose we take the sets of subsets of $mathbb{R}$ and consider the operations $cup(cdot, cdot);,; cap(cdot, cdot);,; cdot^complement;,; mathrm{int}(cdot);,; overline{(cdot)} $ , so union, intersection, complement, interior, and closure.
$cup$ and $cap$ distribute over each other and are duals with respect to $cdot^complement$ and $x^complement{^complement} = x$ .
Interior are closure are also dual and each is idempotent.
$$ mathrm{int}(x^complement)^complement = overline{x} tag{1} $$
$$ mathrm{int}(mathrm{int}(x)) = mathrm{int}(x) tag{2} $$
And the union of the closures is the closure of the (finite) union(s).
$$ overline{x} cup overline{y} = overline{x cup y} tag{3} $$
Here's an example of a finite structure satisfying the rules. The domain of our structure is ${text{-1}, 0, 1}$
$$ x cap y stackrel{df}{=} min(x, y) tag{4} $$
$$ x cup y stackrel{df}{=} max(x, y) tag{5} $$
$$ x ^complement stackrel{df}{=} -x tag{6} $$
$$ mathrm{int}(text{-1}) stackrel{df}{=} text{-1} \
mathrm{int}(0) stackrel{df}{=} 0 \
mathrm{int}(1) stackrel{df}{=} 0 tag{7} $$
$$ overline{(text{-1})} stackrel{df}{=} 0 \
overline{0} stackrel{df}{=} 0 \
overline{1} stackrel{df}{=} 1 tag{8} $$
Note that it isn't possible to swap the definitions in (7) and (8) because of (3) ... which guarantees that union and closure "move things in a similar direction".
abstract-algebra
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
$endgroup$
Is there a name for an algebraic structure equipped with an involution and two additional unary functions that "act like" the interior and closure of a set?
I'm after something a little bit like an $i$-lattice / DeMorgan algebra but additionally with something that "feels like" $square$ and $lozenge$ from modal logic.
Suppose we take the sets of subsets of $mathbb{R}$ and consider the operations $cup(cdot, cdot);,; cap(cdot, cdot);,; cdot^complement;,; mathrm{int}(cdot);,; overline{(cdot)} $ , so union, intersection, complement, interior, and closure.
$cup$ and $cap$ distribute over each other and are duals with respect to $cdot^complement$ and $x^complement{^complement} = x$ .
Interior are closure are also dual and each is idempotent.
$$ mathrm{int}(x^complement)^complement = overline{x} tag{1} $$
$$ mathrm{int}(mathrm{int}(x)) = mathrm{int}(x) tag{2} $$
And the union of the closures is the closure of the (finite) union(s).
$$ overline{x} cup overline{y} = overline{x cup y} tag{3} $$
Here's an example of a finite structure satisfying the rules. The domain of our structure is ${text{-1}, 0, 1}$
$$ x cap y stackrel{df}{=} min(x, y) tag{4} $$
$$ x cup y stackrel{df}{=} max(x, y) tag{5} $$
$$ x ^complement stackrel{df}{=} -x tag{6} $$
$$ mathrm{int}(text{-1}) stackrel{df}{=} text{-1} \
mathrm{int}(0) stackrel{df}{=} 0 \
mathrm{int}(1) stackrel{df}{=} 0 tag{7} $$
$$ overline{(text{-1})} stackrel{df}{=} 0 \
overline{0} stackrel{df}{=} 0 \
overline{1} stackrel{df}{=} 1 tag{8} $$
Note that it isn't possible to swap the definitions in (7) and (8) because of (3) ... which guarantees that union and closure "move things in a similar direction".
abstract-algebra
abstract-algebra
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
asked Dec 3 '18 at 0:42
Gregory NisbetGregory Nisbet
566312
566312
$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22
add a comment |
$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22
$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22
add a comment |
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$begingroup$
I think this falls into the category of Boolean algebras with operators?
$endgroup$
– Noah Schweber
Dec 3 '18 at 1:14
$begingroup$
@NoahSchweber Isn't it amazing that one could own that book for $389?!!
$endgroup$
– rschwieb
Dec 3 '18 at 2:22