Re: countable Heyting algebras (third try)

Re: countable Heyting algebras (third try)

[Dana Scott]

On Nov 26, 2007, at 10:09 AM, Dana Scott wrote,
recall in an earlier posting:

> Now, F can also be considered as the clopens of the Cantor space
> 2^N.  The cHa of all opens is fully non-atomic, but it is uncountable.
> Call it C for Cantor.  The not-not stable elements of C are the so-
> called regular open sets.  They form an uncountable cBa which is the
> completion of F.  But we want to ask if there are there interesting
> countable subalgebras of C?  We note that elements of C are determined
> by the elements of F they contain.  (In fact, the cHa C is isomorphic
> to the lattice of ideals of F.)

The phrase "fully non-atomic" was not well chosen.  Let us use
"atomless"
to mean "no minimal non-zero elements", and let us say a Ha is "gapless"
if we cannot have elements a < b with [a, b] = {a, b} (i.e. nothing
strictly between).  An atomless Ba is always gapless.

Now the cHa C above, as I pointed out in the previous message, DOES
have gaps, even though it is atomless.  (The Ba F is atomless, and
it generates C by taking unions of clopen sets to make opens.  Thus,
no open set could be an atom in C.)

Take any point t in the Cantor set 2^N.  Let b = 2^N and let
a = 2^N \ {t}.  Clearly, a is a dense open set and [a, b] is a gap.
By removing one pont at a time, we can have a whole sequence of
dense open sets a_0 < a_1 < ... < a_n with each [a_i, a_(i+1)]
being a gap.  Note that negation in C gives --a_i = b, since
in topological lattices double negation is interior-of-closure.

In general, in any Ha which F generates, if a < b and [a, b] is a
gap, then b =< --a.  Because if not, then b /\ -a is non-zero.  By
the generation, there must be a non-zero e in F with e =< b /\ -a.
Thus, e /\ a = 0.  Because F is atomless, we can write e = f \/ g,
with two disjoint, non-zero elements of F.  But then c = a \/ f
is an element strictly between a and b.

This comment shows that gaps, if they exist are somewhat limited.
But, C has many gaps, and in general an interval [a, --a] might
be quite large.  Remember, assuming a = --a for all a makes
the Ha Boolean.

So, I have not really made much progress in answering how F
might generate a countable, non-boolean Ha.  I am guessing there
are many non-isomorphic ways this can happen.