Question

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(10 pt) For a partially ordered set (S, <) a "least element" is an element a e S for which a x for all x e S. In other words, a is "less than or equal to" all elements in the set In contrast, a "minimal element" is an element m € S for which·m → x m for all x E S. In other words, no element in the set is "less than or equal to" m (besides m itself). 6. a. Show that if a partially ordered set has a least element then it must be unique b. Show by example that it is possible for a partially ordered set to have multiple distinct minimal elements. 2,7, 7. (10 pt) Identify each of the following graphs as a cycle (C), wheel (W), complete graph (K) and/or bipartite graph (B). Circle your answers. C. d. e.

Answer 1

6)

a) Let a be least element of poset.

To prove: It is unique.

On contradiction, Let b be another least element.

Since a is least element

  $\Rightarrow a\leq b$

Also,   (anti symetric property)

b< a

Thus, a=b which proves uniqueness.

b) see attached image.

7.

Lets have definition of terms used-

Bipartite Graph- If we are able to find two disjoint subsets U and V of vertices of graph such that each edge of graph connect one vertices of U to one vertices of V.

Cycle-If connected graph contains one cycle. In this number of vertices are same as number of edeges.

Wheel-If there exist a vertex which is connected to all other vertex.

Complete Graph-If every vertices are connected to each other.

For solution see attached image.
6-b) , 133 ae minimaul elem entS 2 1 17 113 is least element

a) Not wheel as it has no such veytex which is connecied to all othas vestex Also not Bi-pcstite b) V-a,e,f C. Not Wheel By cla liniton Not complee.> as aleb not connected C) Not cycle →as numbes of vextices + numbes of edges No wheel→ By definition Not complete-s as aed not co nucted Not Bipox tite→ as we cannot get tuoo disjoint sels to sctlsfy clo fnitlon

d) Not uyneel → By definition e. -> as numbes of vextices num be t Edges Not complete as ak d not connacte Two cisioint sets Bipastite cl cu Not c닝cle→ (as same 匹ea so n as above wheel→ a is veztex which is co nn ect ecl with au othex vestex Complete- By de ni tion Not Bi paštite → Not able to get tuo dlisjoint ubset which follous desinition