# Boolean lattice - Interval sublattice

I cannot conclude with the demonstration of the following exercise, I have already verified that the map is a homomorphism, but I do not know how to calculate $$f(L)$$.

For a Boolean lattice $$B$$ and $$a, b \in B$$ such that $$a \leqslant b$$, show that the interval sublattice $$\displaystyle [a,b] := \uparrow \negmedspace a \cap \downarrow \negmedspace b = \{ x \in B | a \leqslant x \leqslant b \}$$ is a Boolean lattice. [Hint. First show that for any distributive lattice $$L$$ the map $$f:L \rightarrow L$$, given by $$f(x) := (x \lor a) \land b$$, is a homomorphism. Then calculate $$f(L)$$.]

• Well, do you have a guess for what you want $f(L)$ to be? What would you want it to be in order to solve the problem? – Eric Wofsey Dec 16 '19 at 0:17

Given that $$a \leq b$$, by distributivity we get $$f(x) = (x \vee a) \wedge b = (x \wedge b) \vee a,$$ whence $$f(L) \subseteq [a,b]$$.

For $$x \in [a,b]$$, $$f(x)=(x\vee a) \wedge b = x \wedge b = x,$$ and thus, $$f(L)=[a,b]$$.

Since $$L$$ is Boolean, in particular for $$x \in [a,b]$$, $$f(x) \wedge f(x') = f(x \wedge x') = f(0) = a,$$ and likewise $$f(x) \vee f(x') = b$$.

Hence $$[a,b]$$ is Boolean.

One doesn't "calculate" $$f(L)$$, one makes a reasonable guess (eg $$f(L) \stackrel{?}{=} [a,b]$$ ) and then one tries to establish the guessed equality.

$$f(L) \subseteq [a,b]$$ :

Let $$x \in L$$. Clearly, $$f(x) \leqslant b$$. We show that $$f(x)\geqslant a$$ :
Since $$a\leqslant a\vee x$$ and $$a\leqslant b$$, $$a$$ is a lower bound of $$\{a\vee x,\, b\}$$. Hence, $$a \leqslant \min \{a\vee x,\, b\} = f(x)$$.

Can you show the other inclusion? If no, tell me where you're stuck.