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What is the covering relation of the partial ordering {(A, B) | A ⊆ B} on the power set of S, where S = {a, b, c}?

i’m getting

R={(Ф, {a}), (Ф, {b}), (Ф, {c}), (Ф, {a, b}), (Ф, {b, c}), (Ф, {a, c}), (Ф, {a, b, c}), ({a}, {a, b}), ({a}, {a, c}), ({b}, {b, c}), ({b}, {a, b}), ({c}, {b, c}), ({a}, {a, b}), ({a}, {a, b, c}), ({b}, {a, b, c}), ({c}, {a, b, c}), ({a, b}, {a, b, c}), ({b, c}, {a, b, c}), ({a, c}, {a, b, c}) }


but in Rosen answer gives is

(∅, {a}), (∅, {b}), (∅, {c}), ({a}, {a, b}), ({a}, {a, c}), ({b}, {a, b}), ({b}, {b, c}), ({c}, {a, c}), ({c}, {b, c}), ({a, b}, {a, b, c}), ({a, c}, {a, b, c})({b, c}, {a, b, c})
in Set Theory & Algebra by Active (5.1k points) | 67 views
+1
according to definition,
subset B covers subset A when you add only one element to subset A which is not in A.
if you write ( $\phi$, {a, b}) then $\phi \subseteq \{a\} \subseteq \{a, b\} $ or $ \phi \subseteq \{b\} \subseteq \{a, b\} $ which is violating the definition of covering relation.
So,  for every pair, add only one element in B which is not in A (or)  make the hasse diagram and then check.
0
can u pls refer the definition of covering relations
I didn't find it in Rosen
+1
thank u

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