DENSE SUBSETS OF PARTIALLY ORDERED SETS
Alexander Abian
The existence of generic filters for Cohen's extension of the minimal
model and conditions imposed on the amount of dense subsets in Martin's
Axiom is due to the fact which is stated in the Theorem below.
In what follows <* stands for "less than or equal"
We recall that a subset D of a partially ordered (P,<*) is
called "DENSE" iff for every element x of P there is an
element y of D such that y <* x.
THEOREM (Abian). Any partially order set (P,<* ) has either
finitely many or else at least continuum many dense subsets.
PROOF. Two cases:
(i) (P,<* ) has a strictly decreasing sequence such as
(1) p(0) > p(1) > p(2) > p(3) > ...> p(n) > .....
Clearly, P has continuum many strictly decreasing subsequences S(i)
of the sequence (1). But then obviously for each S(i),
S(i) union P - S(i) is a dense subset of P
Thus, P has at least continuum many dense subsets.
(ii) (P,<* ) has no strictly decreasing sequence . So any strictly
decreasing nonempty sequence is finite and produces a minimal element
of P. Let M be the set of all minimal elements of P. Clearly, M
must be a subset of any dense subset of P. Now, if P - M is
finite then P has finitely many dense subsets. If P - M is infinite
than since
M union any subsets of P - M is a dense subset of P
P has at least continuum many dense subsets. QED.
P.S. So there is no partially ordered set which has a countable infinite
number of dense subsets.
--
-------------------------------------------------------------------------
ABIAN TIME-MASS EQUIVALENCE FORMULA T = A m^2 in Abian units.
ALTER EARTH'S ORBIT AND TILT TO STOP GLOBAL DISASTERS AND EPIDEMICS.
JOLT THE MOON TO JOLT THE EARTH INTO A SANER ORBIT.ALTER THE SOLAR SYSTEM.
REORBIT VENUS INTO A NEAR EARTH-LIKE ORBIT TO CREATE A BORN AGAIN EARTH(1990)
THERE WAS A BIG SUCK AND DILUTION OF PRIMEVAL MASS INTO THE VOID OF SPACE
Jeremy Boden
<ab...@iastate.edu> writes
The dense bit is interesting though.
--
Jeremy Boden mailto:jer...@jboden.demon.co.uk
Alexander Abian
Correction of a typos
in (1) " S = i" s added
in (1*) S insted of S(i)
The existence of generic filters for Cohen's extension of the minimal
model and conditions imposed on the amount of dense subsets in Martin's
Axiom is due to the fact which is stated in the Theorem below.
In what follows <* stands for "less than or equal"
We recall that a subset D of a partially ordered (P,<*) is
called "DENSE" iff for every element x of P there is an
element y of D such that y <* x.
THEOREM (Abian). Any partially order set (P,<* ) has either
finitely many or else at least continuum many dense subsets.
PROOF. Two cases:
(i) (P,<* ) has a strictly decreasing sequence such as
(1) S = p(0) > p(1) > p(2) > p(3) > ...> p(n) > .....
Clearly, S has continuum many strictly decreasing subsequences S(i)
of the sequence (1). But then obviously for each S(i),
(1*) S(i) union P - S is a dense subset of P
Thus, P has at least continuum many dense subsets.
(ii) (P,<* ) has no strictly decreasing sequence . So any strictly
decreasing nonempty sequence is finite and produces a minimal element
of P. Let M be the set of all minimal elements of P. Clearly, M
must be a subset of any dense subset of P. Now, if P - M is
finite then P has finitely many dense subsets. If P - M is infinite
than since
M union any subsets of P - M is a dense subset of P
P has at least continuum many dense subsets. QED.
P.S. So there is no partially ordered set which has a countable infinite
number of dense subsets.
--
Alexander Abian
Rephrasing the first 3 lines of the PROOF.
The existence of generic filters for Cohen's extension of the minimal
model and conditions imposed on the amount of dense subsets in Martin's
Axiom is due to the fact which is stated in the Theorem below.
In what follows <* stands for "less than or equal"
We recall that a subset D of a partially ordered (P,<*) is
called "DENSE" iff for every element x of P there is an
element y of D such that y <* x.
THEOREM (Abian). Any partially order set (P,<* ) has either
finitely many or else at least continuum many dense subsets.
PROOF. Two cases:
(i) (P,<* ) has a strictly decreasing sequence
p(0) > p(1) > p(2) > p(3) > ...> p(n) > .....
Let
(1) S = { p(0), p(1), p(2), p(3),..., p(n) ... }
Clearly, S has continuum many strictly decreasing subsequences S(i)
But then obviously for each S(i),
(1*) S(i) union (P - S) is a dense subset of P
Thus, P has at least continuum many dense subsets.
(ii) (P,<* ) has no strictly decreasing sequence . So any strictly
decreasing nonempty sequence is finite and produces a minimal element
of P. Let M be the set of all minimal elements of P. Clearly, M
must be a subset of any dense subset of P. Now, if P - M is
finite then P has finitely many dense subsets. If P - M is infinite
then since