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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 views

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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 $B$ …
Joel David Hamkins's user avatar
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 $p$. …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar
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 …
Joel David Hamkins's user avatar

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