1. **Problem Statement:** Construct the Hasse diagram for the set $S=\{1,3,9,15,45,135\}$ with the relation defined by divisibility ($x \mid y$ means $x$ divides $y$). Also, determine if $S$ forms a lattice under this relation. 2. **Understanding the Relation:** The relation $x \mid y$ means $x$ divides $y$ without remainder. This is a partial order on $S$ because it is reflexive, antisymmetric, and transitive. 3. **Step 1: List all divisibility relations in $S$:** - $1$ divides all elements. - $3$ divides $9, 15, 45, 135$. - $9$ divides $45, 135$. - $15$ divides $45, 135$. - $45$ divides $135$. - $135$ divides itself only. 4. **Step 2: Identify covering relations for the Hasse diagram:** A covering relation $x \lessdot y$ means $x \mid y$ and there is no $z$ such that $x \mid z \mid y$ with $z \neq x,y$. - $1 \lessdot 3$ and $1 \lessdot 15$ (since no element between 1 and 3 or 15) - $3 \lessdot 9$ and $3 \lessdot 15$ (note $3 \mid 15$ directly) - $9 \lessdot 45$ - $15 \lessdot 45$ - $45 \lessdot 135$ 5. **Step 3: Draw the Hasse diagram:** - Bottom element: $1$ - Above $1$: $3$ and $15$ - Above $3$: $9$ - Above $9$ and $15$: $45$ - Top element: $135$ 6. **Step 4: Check if $S$ is a lattice:** A lattice requires every pair of elements to have a unique least upper bound (join) and greatest lower bound (meet). - For example, consider $9$ and $15$: - Common upper bounds: $45, 135$ - Least upper bound (join): $45$ - Common lower bounds: $1, 3$ - Greatest lower bound (meet): $3$ - Similarly, all pairs have unique join and meet. **Conclusion:** $S$ with divisibility is a lattice. **Final answer:** The Hasse diagram is constructed as described, and $S$ is a lattice under divisibility.