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Next, we can look at the composite functions:

\[g \circ f: 2^X \rightarrow 2^X\] \[f \circ g: 2^Y \rightarrow 2^Y\]

Comments

  • 1.

    Given that \(f\) and \(g\) form a Galois connection, it follows that these composites are closure operators.

    For a function T mapping a partially ordered set into itself, this means that T is:

    • Monotone: \(a \subseteq b \implies T(a) \subseteq T(b)\)
    Comment Source:Given that \\(f\\) and \\(g\\) form a Galois connection, it follows that these composites are **closure operators**. For a function T mapping a partially ordered set into itself, this means that T is: * Monotone: \\(a \subseteq b \implies T(a) \subseteq T(b)\\) *
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