Reputation:
Prove by induction.Every partial order on a nonempty finite set at least one minimal element.
How can I solve that question ?
Upvotes: 0
Views: 1635
Reputation: 1597
If the partial order has size 1, it is obvious.
Assume it is true for partial orders <n
, and then take a partial order (P,<)
has size n
.
Pick x
in P
. Let P(<x) = { y in P : y<x }
If P(<x)
is empty, then x
is a minimal element.
Otherwise, P(<x)
is strictly smaller than P
, since x
is not in P(<x)
. So the poset (P(<x),<)
must have a minimal element, y
.
This y
must be a minimal element of P
since, if z<y
in P
, then z<x
, and hence z
would be in P(<x)
and smaller than y
, which contradicts the assumption that y
was minimal in P(<x)
.
Upvotes: 0
Reputation: 11849
It is trivially true if there is only one element in the poset. Now suppose it is true for all sets of size < n. Compare the nth element to the minimal element of the (n-1) poset, which we know to exist. It will either be the new minimal or not or incomparable. It doesn't matter either way. (Why?)
Upvotes: 2