These derivation operators have some nice order-theoretic structure.

First, note the powersets \\(2^X\\) and \\(2^Y\\) are of course partially ordered by the inclusion relation.

Next, it is not hard to see that \\(f: 2^X \rightarrow 2^Y\\) and \\(g: 2^Y \rightarrow 2^X\\) are _order reversing_ functions.

First consider \\(f: 2^X \rightarrow 2^Y\\).

What happens if you apply \\(f\\) to a subset \\(x_1 \subseteq x \subseteq X\\)?

Then, since the \\(x_1\\) is less than or equal to \\(x\\), the set of attributes which apply to every object in \\(x_1\\) must be greater than or equal to the set of attributes which apply to every object in \\(x\\). We weakened the condition by choosing a subset \\(x_1\\) of \\(x\\), which can only increase the set of common attributes.

So \\(x_1 \subseteq x \implies f(x_1) \supseteq f(x)\\), which says that \\(f\\) is order-reversing.

Similarly, decreasing a set of attributes \\(y\\) can only increase the set of objects \\(g(y)\\) which have all the attributes in \\(y\\).

So \\(g\\) is order-reversing too.

First, note the powersets \\(2^X\\) and \\(2^Y\\) are of course partially ordered by the inclusion relation.

Next, it is not hard to see that \\(f: 2^X \rightarrow 2^Y\\) and \\(g: 2^Y \rightarrow 2^X\\) are _order reversing_ functions.

First consider \\(f: 2^X \rightarrow 2^Y\\).

What happens if you apply \\(f\\) to a subset \\(x_1 \subseteq x \subseteq X\\)?

Then, since the \\(x_1\\) is less than or equal to \\(x\\), the set of attributes which apply to every object in \\(x_1\\) must be greater than or equal to the set of attributes which apply to every object in \\(x\\). We weakened the condition by choosing a subset \\(x_1\\) of \\(x\\), which can only increase the set of common attributes.

So \\(x_1 \subseteq x \implies f(x_1) \supseteq f(x)\\), which says that \\(f\\) is order-reversing.

Similarly, decreasing a set of attributes \\(y\\) can only increase the set of objects \\(g(y)\\) which have all the attributes in \\(y\\).

So \\(g\\) is order-reversing too.