There's a few observations we can make just by looking at the lattice:

• If we take any two subsets on the same level, such as \(\{\{a\}, \{b\}\}\) then two move up a level we take the union of the sets, so: \[ \{a\} \cup \{b\} \to \{a, b\} \] Conversly if we wish to move down a level we take the intersection of the two sets, so for the same pair \[ \{a\} \cap \{b\} \to \{\} \]
• Each level expresses the different powersets of \(X\) with a given cardinality. So the bottom level is the powerset of cardinality 0 (i.e. the empty set), the next level up is the powerset of cardinality 1 (i.e. the singletons), then cardinality 2 and finally cardinality 3 at the top level which is just the set \(X\) itself.
• Naturally as we move up the lattice the sets get larger (with respect to inclusion) and as we move down the lattice the sets get smaller.

Here the lower bound is any element \((\sqcup S) \in L\) in the larger lattice in which the following property holds:

In our case we know that \(\leq \ \equiv \ \subseteq \), furthermore we have that the elements of \(S\) are the two sets \(\{a\}\) and \(\{a, b\}\), hence we can rephrase this condition as

Clearly the lower bound in this instance is then just:

The greatest lower bound is simply the intersection of all lower bounds hence:

The upper bound is just the flipped condition of the lower bound, i.e.:

The sets which satisfy this condition are:

The least upper bound is the smallest of the sets which form the upper bound, i.e.: