Proving Boolean Lattice Complementarity in [a,b]

[Aryth1]

The problem is this:

Let $L$ be a Boolean lattice. For all $a<b\in L$, prove that the interval $[a,b] = \uparrow a \cap \downarrow b$ is a Boolean lattice under the partial ordering inherited from $L$.

What I've managed to do so far:

I used the fact that $[a,b]$ was a poset and showed that, for any $x,y\in [a,b]$, $x\vee y\in [a,b]$ and $x\wedge y\in [a,b]$ which proved that $[a,b]$ was a sublattice of $L$. From this we get that $[a,b]$ is distributive for free, since if $L$ had a sublattice that wasn't distributive, then $L$ would not be distributive. All that remains is to show that $[a,b]$ is complemented. I cannot seem to figure this out. I know that, if we take some $x\in [a,b]$ then a complement exists in $L$ since $L$ is a Boolean lattice, but I don't know how to show that that complement is in $[a,b]$ also.

Any hints would be greatly appreciated!

 

[Aryth1]

Nevermind, I figured it out. Close or delete if you like.

 

[MarkFL]

Nevermind, I figured it out. Close or delete if you like.

If you want, and it is entirely up to you, you could post your solution for the benefit of others who might gain from your solution. :D

 

[Aryth1]

If you want, and it is entirely up to you, you could post your solution for the benefit of others who might gain from your solution. :D

Ah, OK then.

I'll limit the problem to proving that $[a,b]$ is complemented.

Let $x\in [a,b]$. Then, since $L$ is a Boolean lattice, $x$ has a complement in $L$, call it $y$. We claim that the complement of $x$ in $[a,b]$ is $z = (y\vee a)\wedge b$. To see this, observe that:

$x\wedge z = x\wedge (y\vee a)\wedge b = [(x\wedge y)\vee (x\wedge a)]\wedge b = (\bot\vee a)\wedge b = a$​

Similarly, $x\vee z = b$ and $z$ suffices as a complement for $x$ in $[a,b]$. Since $x$ was arbitrary, $[a,b]$ is complemented.

 

[blue_raver22]

First of all, great job on proving that $[a,b]$ is a sublattice of $L$ and that it is distributive. To show that $[a,b]$ is complemented, we need to show that for every $x\in [a,b]$, there exists a unique element $x^*\in [a,b]$ such that $x\vee x^* = b$ and $x\wedge x^* = a$.

To start, let's consider the set $S = \{x^*\in L | x\vee x^* = b \text{ and } x\wedge x^* = a\}$. We want to show that $x^*$ exists and is unique.

First, let's show that $x^*$ exists. Since $b\in L$, we know that $x\vee b = b$. This means that $b$ is an upper bound for $x$ in $L$. Similarly, $a$ is a lower bound for $x$ in $L$. Since $L$ is a Boolean lattice, we can take the supremum of all upper bounds of $x$ and the infimum of all lower bounds of $x$ to get the unique element $x^*\in L$ such that $x\vee x^* = b$ and $x\wedge x^* = a$. Therefore, $x^*\in S$ and $x^*$ exists.

Next, we need to show that $x^*$ is unique. Suppose there exists another element $y^*\in S$ such that $x\vee y^* = b$ and $x\wedge y^* = a$. Then, we have $x\vee x^* = b = x\vee y^*$. Since $L$ is a lattice, this means that $x^* = y^*$. Similarly, we have $x\wedge x^* = a = x\wedge y^*$, which again implies that $x^* = y^*$. Therefore, $x^*$ is unique.

Hence, we have shown that for every $x\in [a,b]$, there exists a unique element $x^*\in [a,b]$ such that $x\vee x^* = b$ and $x