Another Boolean Lattice Problem

[Aryth1]

My problem is this:

Let $L$ be a Boolean lattice. Prove that $L$ is atomic if and only if its order dual $L^{op}$ is atomic.

My proof is going like this so far:

Let $L$ be a Boolean lattice. Suppose first that $L$ is atomic. Then, by definition, every lowerset in $L$ contains an atom. Since $L$ is a Boolean lattice, each of these atoms have a complement. These complements serve as co-atoms for $L$ by lemma (I proved this as a lemma for this problem), i.e. as atoms in $L^{op}$.

What I can't seem to figure out is how to show that every lowerset in $L^{op}$ contains a co-atom from $L$. Any help is appreciated.

 

[Deveno]

I would think that this follows from:

$(L^{\text{op}})^{\text{op}} = L$.

 

[Aryth1]

I would think that this follows from:

$(L^{\text{op}})^{\text{op}} = L$.

I'm not really sure I follow. I mean, I get your equality, that's certainly true. I keep having trouble. The only definition of atomic I have for Boolean lattices so far is that every lower set in the lattice contains an atom. So, if I assume that $L$ is an atomic Boolean lattice, I can assume that all the lower sets contain atoms in $L$. By duality, this only means that all the upper sets of $L^{op}$ contain co-atoms of $L^{op}$. I have nothing so far that says anything about the lower sets of $L^{op}$.

I have boiled the problem down to this: If every element in $L$ is above an atom of $L$ (given by atomicity), then every element in $L$ must be below a co-atom of $L$.

I'm not necessarily sure I can show this with what I know (I could be missing something). Here's what I've done so far. I let $y\in L$. Since $L$ is a Boolean lattice, there exists an atom $a\in L$ such that $a \leq y^c$. I guess I need to show that this means that $y\leq a^c$, right? Because then, since $y$ is in some upper set of $L$, then $a^c$ (which is an atom in $L^{op}$) is in that upper set as well and $L^{op}$ must be atomic.

The reverse implication is true by duality, replacing $L$ with $L^{op}$ and vice versa.

EDIT: I wonder if this is OK.

$a\leq y^c \iff a\wedge y = \bot \iff a\wedge y \leq \bot \iff y\leq \bot \vee a^c = a^c$

Then $y\leq a^c$ as desired. We let $\uparrow x$ be the upper set containing $y$. This would then show that every arbitrary lower set in $L^{op}$ contains an atom of $L^{op}$ as desired.

 

[Deveno]

This was my thought:

If $L$ atomic implies $L^{\text{op}}$ atomic, then the same proof (applied to $L^{\text{op}}$) shows that:

$(L^{\text{op}})^{\text{op}}$ is atomic, but this is $L$.

Let's look at an actual atomic lattice, to see how this works:

define $L = \mathbf{2}^X$, where $X = \{a,b,c\}$ (the power set of a 3-element set, partially ordered by inclusion).

The atoms of $L$ are $\{a\},\{b\},\{c\}$. Let's call these elements $A,B,C$.

The co-atoms of $L$ are:

$\{b,c\}, \{a,c\},\{a,b\} = A^c,B^c,C^c$. <---these are the atoms of $L^{\text{op}}$.

As you can see, a co-atom is just $1 \wedge Y$, where $Y$ is an atom (here $\wedge$ is the "meet" or "greatest lower bound" of two elements of $L$).

I think you are being confused by duality. A co-atom in $L$ is an ATOM in $L^{\text{op}}$.

 

[Aryth1]

This was my thought:

If $L$ atomic implies $L^{\text{op}}$ atomic, then the same proof (applied to $L^{\text{op}}$) shows that:

$(L^{\text{op}})^{\text{op}}$ is atomic, but this is $L$.

Let's look at an actual atomic lattice, to see how this works:

define $L = \mathbf{2}^X$, where $X = \{a,b,c\}$ (the power set of a 3-element set, partially ordered by inclusion).

The atoms of $L$ are $\{a\},\{b\},\{c\}$. Let's call these elements $A,B,C$.

The co-atoms of $L$ are:

$\{b,c\}, \{a,c\},\{a,b\} = A^c,B^c,C^c$. <---these are the atoms of $L^{\text{op}}$.

As you can see, a co-atom is just $1 \wedge Y$, where $Y$ is an atom (here $\wedge$ is the "meet" or "greatest lower bound" of two elements of $L$).

I think you are being confused by duality. A co-atom in $L$ is an ATOM in $L^{\text{op}}$.

I still don't follow. I have a Boolean lattice, $L$, to work with. I assume $L$ is atomic... But I have to prove that $L^{op}$ is atomic. If I assume that $L^{op}$ is atomic first, I then can't say anything about $L$ itself except that it's a Boolean lattice until I prove that it is also atomic. I'm not sure how what you have leads me to the result... Unfortunately. I am slightly familiar with what you're doing though. I did prove later that a complete Boolean lattice is atomic if and only if it is order isomorphic to the power set of some set, but I can't use this for my result.

I know that a co-atom in $L$ is an atom in $L^{op}$, however, I can do the entire proof in $L$ alone by showing that every element must be less than a co-atom of $L$ (or, dually, greater than an atom in $L^{op}$) provided it is greater than an atom of $L$. That would, indeed, prove that $L^{op}$ is atomic since every element is less than a co-atom in $L$ and so, dually, every element in $L^{op}$ is greater than an atom of $L^{op}$. The problem is that I'm quite sure how to do it.

 

[Deveno]

Ok, you are given that $L$ is atomic.

So, given $b > 0 \in L$, we have an atom $a$ with: $0 < a \leq b$.

This means that we have $b^c \leq a^c < 1 (= 0^c)$, since $L$ is Boolean.

This is true for EVERY $b > 0$ of $L$, including $b = x^c$ for any $x$, so if $x < 1$, there is a co-atom $y$ with:

$x \leq y < 1$, namely $y = (a')^c$, where $a'$ is the atom for which $0 < a' \leq x^c$.

Since this is true for ANY $x < 1$, this establishes that $L$ is "co-atomic", i.e., that $L^{\text{op}}$ is atomic

(All we do to $L$ to get $L^{\text{op}}$ is turn $L$ "upside-down", formally replacing all $<$ with $>$ and all $\leq$ with $\geq$).

The key to this is recognizing the map $b \mapsto b^c$ is surjective, because we have any $x \in L$ is the complement of $x^c$. It's also injective, as well:

If $b^c = d^c$, then surely $b = (b^c)^c = (d^c)^c = d$.

This is where the Boolean nature of $L$ makes itself felt, we need $L$ to be a complemented lattice.

Note the mapping $b \mapsto b^c$ is order-REVERSING on $L$, and order-PRESERVING from $L \to L^{\text{op}}$.

Duality means: $L$ and $L^{\text{op}}$ are isomorphic lattices. This is not true of lattices in general, imagine a lattice whose diagram forms the letter "Y" (with 0 at the "tail"). For this lattice, the "opposite lattice" has no 0. So our original lattice is atomic (the sole atom being where the three lines of the "Y" meet), but it's opposite is not (atomic lattices must possesses a 0).

 

[Aryth1]

Ok, you are given that $L$ is atomic.

So, given $b > 0 \in L$, we have an atom $a$ with: $0 < a \leq b$.

This means that we have $b^c \leq a^c < 1 (= 0^c)$, since $L$ is Boolean.

This is true for EVERY $b > 0$ of $L$, including $b = x^c$ for any $x$, so if $x < 1$, there is a co-atom $y$ with:

$x \leq y < 1$, namely $y = (a')^c$, where $a'$ is the atom for which $0 < a' \leq x^c$.

Since this is true for ANY $x < 1$, this establishes that $L$ is "co-atomic", i.e., that $L^{\text{op}}$ is atomic

(All we do to $L$ to get $L^{\text{op}}$ is turn $L$ "upside-down", formally replacing all $<$ with $>$ and all $\leq$ with $\geq$).

The key to this is recognizing the map $b \mapsto b^c$ is surjective, because we have any $x \in L$ is the complement of $x^c$. It's also injective, as well:

If $b^c = d^c$, then surely $b = (b^c)^c = (d^c)^c = d$.

This is where the Boolean nature of $L$ makes itself felt, we need $L$ to be a complemented lattice.

Note the mapping $b \mapsto b^c$ is order-REVERSING on $L$, and order-PRESERVING from $L \to L^{\text{op}}$.

Duality means: $L$ and $L^{\text{op}}$ are isomorphic lattices. This is not true of lattices in general, imagine a lattice whose diagram forms the letter "Y" (with 0 at the "tail"). For this lattice, the "opposite lattice" has no 0. So our original lattice is atomic (the sole atom being where the three lines of the "Y" meet), but it's opposite is not (atomic lattices must possesses a 0).

Thank you for sticking with me! This is what I was trying to do. My professor gave me a very small hint and I can see it used in here... But I'm not sure how he expected me to get there. I'm still kind of new to this, so thanks also for the details at the end, the information is very helpful and much appreciated.