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The theory of lattices in the sense of order theory. For the number-theoretic notion, use the tag "lattices" instead.
82
votes
5
answers
6k
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How do the compact Hausdorff topologies sit in the lattice of all topologies on a set?
This question is about the space of all topologies on a
fixed set X. We may order the topologies by refinement, so
that τ ≤ σ just in case every τ open set is open in σ.
Equivalently, we say in this c …
18
votes
Accepted
Are these two quotients of $\omega^\omega$ isomorphic?
Very nice question!
They are not isomorphic.
What I claim is that when we take the quotient with respect to density, there is a countably infinite antichain above $0$ having a minimal upper bound, b …
17
votes
Knaster Tarski theorem, example needed
You can use this to prove the Cantor-Schröder-Bernstein theorem, which asserts that whenever $A$ injects into $B$ and $B$ injects into $A$, then they are bijective. Namely, suppose that $f:A\to B$ and …
17
votes
Accepted
Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})...
The answer is no.
In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. Namely, one …
13
votes
Accepted
Ultrafilter lemma for arbitrary lattice
It is equivalent to AC.
Consider any collection $A$ of nonempty sets, and let $\newcommand\P{\mathbb{P}}\P$ be the set of partial choice functions, so that $p\in\P$ if and only if $p$ is a partial fun …
11
votes
LUB and GLB on a lexicographically ordered complete lattice product
There are two natural orders to put on the product of two lattices, the product order and the lexical order.
Product order: (a,b) ≤ (a',b') if and only if a ≤ a' and b ≤ b'
Lexical order: (a …
9
votes
Accepted
Pseudocomplements in the lattice of topologies
Yes, and in fact, most familiar topologies do not have a pseudo-complement.
To see this, notice that that it often happens with a topology
$\tau$ on a set $X$ that there are non-open sets $A$ and $B$ …
6
votes
Accepted
Is every finite poset a subset of a finite complemented distributive lattice?
As Sam mentioned in the comments, the answer to question 1 is yes. One can map every condition $p$ in the partial order to the lower cone, the set $S_p=\{q\in P\mid q\leq p\}$ of conditions below $p$. …
5
votes
Accepted
Does the collection of coverings on a set $X$ form a lattice when ordered by refinement?
If there are only finitely many points, it is a lattice, and it is always an upper semi-lattice. But in the infinite case, it is not a lattice.
François's comment on Fedor's answer shows that is it a …
5
votes
Is every lattice the fixed-point set of an order-preserving endomorphism of ⋄^n?
For all finite lattices, the answer is Yes.
More generally, for all complete lattices, the answer is Yes, and for all incompleteness lattices, the answer is No.
(Complete = every set has a LUB and …
5
votes
Accepted
Semilattices in atomless boolean algebras
The answer to the original question is that no, in fact we can never do this.
Theorem. No nontrivial Boolean algebra has a cofinal
subset of B-{1} that is a join-semilattice. Indeed, B cannot have a …
5
votes
Complete De Morgan algebra
I claim that the property is true in every de Morgan algebra,
whenever the expressions in it make sense (on either side). The issue about making
sense is that when $I$ is infinite, the expression $\bi …
4
votes
Accepted
getting one tower from two
This is a fantastic question! I spent the whole morning thinking
about it, and I finally have a solution.
The answer is no, not necessarily.
To build a counterexample, I claim first that there is a …
4
votes
Accepted
Lattice of differences between ultrafilters
I've got it!
Theorem. The lattices of the form
$D(U,V)$ admit a complete classification by the isomorphism classes of $U$ and $V$ and the question of whether $U\neq V$.
The point is that the latti …
4
votes
Countable atomless boolean algebra covered by a larger boolean algebra
Edit: This is the answer to the original question. See my other answer for the answer to the revised question.
The answer is that every such f is an isomorphism. Thus, there is such an f only when …