Difference between revisions of "Disjoint sum of partially ordered sets"
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− | <TR><TD valign="top">[a1]</TD> <TD valign="top"> W.T. Trotter, | + | <TR><TD valign="top">[a1]</TD> <TD valign="top"> W.T. Trotter, "Partially ordered sets" R.L. Graham (ed.) M. Grötschel (ed.) L. Lovász (ed.) , ''Handbook of Combinatorics'' , '''I''' , North-Holland (1995) pp. 433–480 {{ZBL|0841.06001}}</TD></TR> |
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Latest revision as of 06:43, 30 July 2024
2020 Mathematics Subject Classification: Primary: 06A [MSN][ZBL]
disjoint sum of posets
Let $P$ and $Q$ be two partially ordered sets.
The disjoint sum $P+Q$ of $P$ and $Q$ is the disjoint union of the sets $P$ and $Q$ with the original ordering on $P$ and $Q$ and no other comparable pairs. A poset is disconnected if it is (isomorphic to) the disjoint sum of two sub-posets. Otherwise it is connected. The maximal connected sub-posets are called components.
The disjoint sum is the direct sum in the category of posets and order-preserving mappings. The direct product in this category is the Cartesian product $P\times Q$ with partial ordering
$$(p,q)\geq(p',q')\Leftrightarrow p\geq p',q\geq q'.$$
References
[a1] | W.T. Trotter, "Partially ordered sets" R.L. Graham (ed.) M. Grötschel (ed.) L. Lovász (ed.) , Handbook of Combinatorics , I , North-Holland (1995) pp. 433–480 Zbl 0841.06001 |
Disjoint sum of partially ordered sets. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Disjoint_sum_of_partially_ordered_sets&oldid=55926