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# sols6 - M114S and M134 solutions to HW#6 x6.2 Suppose that...

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M114SandM134,solutionstoHW#6 x6.2. Supposethat S isachaininthefunctionspace( A E ).Weneedto findaleastupperboundfor S .Foreach a A ,let S ( a )= { f ( a ) | f S } . Thenweshowthatforeach a A , S ( a )isachainin E . Let f ( a )and g ( a )bothbelongto S ( a ).Since S isachain,either f g or g f inthe pointwiseorder.Eitherway,wehave f ( a ) g ( a )or g ( a ) f ( a ). Foreach a A ,let a =sup S ( a ).Let h : A E begivenby h ( a )= a . Weclaimthat h =sup S . Forall f S , f h becauseforeach a A , f ( a ) S ( a )andhence f ( a ) a .Thisshowsthat h isanupperboundof S inthefunctionspace.Let g beanyupperboundof S ,weshow h g .Todo this,fix a A .Thenas g isanupperboundof S , g ( a ) f ( a )forall f S . So g ( a ) a ,as a =sup S ( a ).So g ( a ) h ( a ).Thisforall a showsthat h =sup S . x6.3. Suppose C P 1 × P 2 isachain,andlet C 1 = { x 1 P 1 | ( x 2 P 2 )[( x 1 ,x 2 ) C ] , C 2 = { x 2 P 2 | ( x 1 P 1 )[( x 1 ,x 2 ) C ] . Claim: C 1 isachainin P 1 . ProofofClaim: If x 1 and x 1 arein C 1 ,then thereexist x 2 and x 2 suchthat( x 1 ,x 2 ) C and( x 1 ,x 2 ) C ;since C isa chain,thesetwopairsmustbecomparable,i.e.,either( x 1 ,x 2 ) ( x 1 ,x 2 )or ( x 1 ,x 2 ) ( x 1 ,x 2 )(withthepartialorderingon P 1 × P 2 ).Nowinthefirstcase wehave x 1 1 x 1 ,andinthesecondcasewehave x 1 1 x 1 .Thiscompletes theproofoftheClaim. Bythesameargumentweshowthat C 2 isachainin P 2 ,andsowecan define M 1 =sup 1 ( C 1 ) , M 2 =sup 2 ( C 2 ) , withthesubscriptsindicatingtheposetsinwhichthesupremaaretaken.Itis nowquiteeasytoverifythat ( M 1 ,M 2 )=sup C, andso C hasaleastupperbound,asrequired.

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