On Isomorphic Lattices I've been recently studying lattices and other ordered structures (after a long break in university) and I'm stuck at a question... tried hard on this one, but all I got was a huge headache.
Question: If $L$ is a lattice and $a,b\in L$, let $X_{a,b}=\{x\in L:x=(x\vee b)\wedge a\}$ and $Y_{a,b}=\{y\in L:y=(y\wedge a)\vee b\}$. Prove that $X_{a,b}$ and $Y_{a,b}$ are isomorphic lattices.
Well, it isn't hard to see that $\{a\wedge b,a\}\subseteq X_{a,b}$ and that $\{b,a\vee b\}\subseteq Y_{a,b}$, but I'm not in position to conclude that there are no other elements in these lattices. If so, the work is done ($X_{a,b}\to Y_{a,b}$, such that $x\mapsto x\vee b$ is a bijection which is a $\vee$-morphism).
 A: Let us see that the order inherited from $L$ gives to
$X_{a,b} = \{ x \in L : x = (x \vee b) \wedge a \}$
the structure of lattice.
Indeed, we will see that $(X_{a,b}, \odot, \oplus)$ is a lattice, where, for $u,v \in X_{a,b}$,
$$u \odot v = ((u \wedge v) \vee b) \wedge a$$
and
$$u \oplus v = ((u \vee v) \vee b) \wedge a.$$
Clearly,
$$u \odot v = ((u \wedge v) \vee b) \wedge a \leq (u \vee b) \wedge a = u,$$
and similarly, $u \odot v \leq v$.
Now suppose that $w \in X_{a,b}$ (so that $w = (w \vee b) \wedge a$) is such that $w \leq u, v$.
If we prove that $w \leq u \odot v$ we can conclude that $u \odot v$ is the meet of $u$ and $v$ in $X_{a,b}$.
Now, $w \leq u \wedge v$ and thus $w \leq (u \wedge v) \vee b$;
since $w \leq a$ (because $w \in X_{a,b}$),
$$w \leq ((u \wedge v) \vee b) \wedge a = u \odot v.$$
The proof that $u \oplus v$ is the join, in $X_{a,b}$, of $u$ and $v$ is analogous.
Hence, $(X_{a,b},\odot,\oplus)$ is a lattice.

Let us now define $\varphi:X_{a,b} \to Y_{a,b}$ and $\psi:Y_{a,b} \to X_{a,b}$ by
$$
\varphi:x \mapsto x \vee b
\quad\text{and}\quad
\psi:y \mapsto y \wedge a.
$$
The maps $\varphi, \psi$ are clearly order-preserving, since they are made from lattice operations which are order-preserving.
If $\varphi(x_1) = \varphi(x_2)$, then $x_1 \vee b = x_2 \vee b$, whence $(x_1 \vee b) \wedge a = (x_2 \vee b) \wedge a$, i.e., $x_1=x_2$.
So $\varphi$ is one-to-one, and analogously $\psi$ is one-to-one.
Moreover,
$$\varphi(\psi(y)) = \varphi(y \wedge a) = (y \wedge a) \vee b = y,$$
and analogously, $\psi(\varphi(x))=x$, for $x \in X_{a,b}$.
So $\varphi$ and $\psi$ are order-preserving and inverse of each other (thus, bijective) and so they are order-isomorphisms.
But we saw that $X_{a,b}$ is a lattice.
So, if $Y_{a,b}$ is order-isomorphic to $X_{a,b}$, then $Y_{a,b}$ is a lattice too, and $\varphi$ and $\psi$ are lattice isomorphisms.
