Is the (strict) level set of a function a sublattice?

Question

I am reading a book having the following statement, which is claimed as correct I do not see why.

Let $f:X\rightarrow Y$ be a function from a lattice $X$ to a chain $Y$. The level set of $f$ is defined as

$\{ x:x\in X,y\leq f(x) \}$

for any $y\in Y$. We know that every level set of $f$ is a sublattice of $X$. Then, the claim is that the following set is also a sublattice of $X$ since $Y$ is a chain:

$\{ x:x\in X,y< f(x) \}$.

If you can help me understand why this claim is correct that would be great.

Answer 1

Let $$A_y=\{x\in X:y\leq f(x)\}$$ and $$B_y=\{x\in X:y<f(x)\}.$$ Note then that $$B_y=\bigcup_{z>y} A_z.$$ Since $Y$ is a chain, this is a union of a chain of nested sets ($z\leq z'$ implies $A_z\supseteq A_{z'}$), and a union of a chain of nested sublattices is a sublattice. Explicitly, for instance, if $x,x'\in B_y$, then there is some $z>y$ such that both $x$ and $x'$ are in $A_z$ (namely, $z=\min(f(x),f(x'))$), so then $x\vee x'$ and $x\wedge x'$ are both in $A_z$ and hence in $B_y$.