I would appreciate some help on proving that an ideal lattice is a sublattice. The question is this:
"Let T be a lattice, and let S be an ideal in T. Show that S is a sublattice of T."
I know that if S is an ideal then with S being a subset of T, x is in S, y is in T, y is less than or equal to x, so y is in S.
I also know that S is a sublattice of T if it has the same meet and join operations as T (and since T is a lattice, it will have a GLB and LUB).
So essentially I have to show that S and T have the same meet and join operations. I'm unsure how to utilize the information I have (from the ideal definition) in order to do that.
Thank you!
Let $x,y\in S,$ let $s\in T$ be their join in $T,$ and let $i\in T$ be their meet in $T.$
Claim 1: $s\in S.$ Because $S$ is an ideal, there exists $z\in S$ such that $x\leq z$ and $y\leq z.$ However, $s$ is characterized in $T$ by the properties that $x\leq s,$ $y\leq s,$ and if $x\leq z$ and $y\leq z,$ then $s\leq z.$ So the second ideal axiom implies that $s\in S.$ You can easily check that because $S\subseteq T,$ $s$ is still the join of $x$ and $y.$ (If not, there is some other element $s'$ such that $x\leq s',$ $y\leq s',$ and $s'\leq s.$ But $s'\in S\implies s'\in T,$ so...)
Claim 2: $i\in S.$ This is immediate from the second axiom of an ideal, because $i\leq x$ and $i\leq y$ by definition of least upper bound. You can similarly check that because $i$ was a least upper bound in $T,$ the same is true in $S.$