Let D be a non-empty set and (D-->D) be the set of all partial functions from D to D. Prove that the partial order ((D--> D), ⊆) (i.e., the set of partial functions ordered by set inclusion) is a complete partial order with bottom.
2026-03-25 17:36:46.1774460206
Prove that the partial order ((D * D), ⊆) is a complete partial order with bottom.
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The empty function is the bottom.
It is not complete.
Let D = {0,1}, f = { (0,1), (1,1) }, g = { (0,0), (1,0) }.
There is no upper bound for {f,g}, nor is there a top.
It is however meet complete.
If F not empty subset D*D, then inf F = $\cap$F.
It is also chain complete.
If K subset D*D is a chain, then sup K = $\cup$K.