Good evening everybody,
I am working on the notion of lattice, which is still very difficult for me to grasp. I have this question:
Suppose that a lattice T can be embedded in a distributive lattice D. Show that T is also distributive.
It is my understanding that an embedding conserves the operations, so if T has ∧,∨, so does D. Then I could verify the distributivity of T by seeing if this is true: x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z)
But then, how am I suppose to do so without any value to verify if it works? Or maybe I'm having a completely wrong angle for this problem? (It's very possible, I'm new to this and completely outside of my field of study.)
Any help would be appreciated,
Thank you
The main idea is: Distributivity can be (theoretically) destroyed by a single counter example. Embedding means injectivity: if $x\neq y$ then $e(x)\neq e(y)$. So if an equation has a counterexample in $T$ it also has a counterexample in $D$.
This can be used formally to describe a direct proof. To avoid unnecessary spoilers, suppose I want to prove that every groupoid (set with a binary operation) that embeds into a semigroup is already a semigroup. So we have two operations $*$ in $T$ and $\cdot$ in $D$. And we know that $D$ is associative. In a first step I write down the associative law in $D$: $$x_D\cdot(y_D \cdot z_D) = (x_D\cdot y_D)\cdot z_D.$$ (This line does not need to occur in the proof, it is only for the construction of the proof).
Now I replace every variable by the embedding of an element of $T$: $$e(x)\cdot\bigl(e(y)\cdot e(z)\bigr)=\bigl(e(x)\cdot e(y)\bigr)\cdot e(z).$$ The embedding $e$ supports the equation $e(x*y) = e(x)\cdot e(y)$. We apply this to the inner parentheses (the outer operation does not support it, yet): $$e(x)\cdot e\bigl(y * z\bigr)=e(x)\cdot\bigl(e(y)\cdot e(z)\bigr)=\bigl(e(x)\cdot e(y)\bigr)\cdot e(z)=e\bigl(x * y\bigr)\cdot e(z)$$
Now, also the outer operations have the structure that allows us to apply the law again: $$e\bigl(x * (y * z)\bigr)=e(x)\cdot e\bigl(y * z\bigr)=e\bigl(x * y\bigr)\cdot e(z)=e\bigl((x * y) *z\bigr)$$
Finally, we apply that $e$ is injective, which implies $$x * (y * z)=(x * y) *z$$ q.e.d.
As you can see it is a very formal process. In the same way you can prove your distributivity law. I chose a different example, here as you obviously need some exercise with this method.