Darío Posted February 29, 2012 Share Posted February 29, 2012 Hi, i should be this proof Let [latex]R[/latex] be an ordering of [latex]A[/latex]. Prove that [latex]R^{-1}[/latex] is also an ordering of [latex]A[/latex], and for [latex]B\subset{A}[/latex], (A) [latex]a[/latex] is the least element of [latex]B[/latex] in [latex]R^{-1}[/latex] if and only if [latex]a[/latex] is the greatest element of [latex]B[/latex] in [latex]R[/latex]. (B) Similarly for (minimal and maximal) and (supremum and infimum. Good look! Link to comment Share on other sites More sharing options...
DrRocket Posted February 29, 2012 Share Posted February 29, 2012 Hi, i should be this proof Let [latex]R[/latex] be an ordering of [latex]A[/latex]. Prove that [latex]R^{-1}[/latex] is also an ordering of [latex]A[/latex], and for [latex]B\subset{A}[/latex], (A) [latex]a[/latex] is the least element of [latex]B[/latex] in [latex]R^{-1}[/latex] if and only if [latex]a[/latex] is the greatest element of [latex]B[/latex] in [latex]R[/latex]. (B) Similarly for (minimal and maximal) and (supremum and infimum. Good look! This is quite clearly a homework problem. You need to show a reasonable attempt before we will help you. Link to comment Share on other sites More sharing options...
Darío Posted March 2, 2012 Author Share Posted March 2, 2012 Yo have reason, in the first i think as follow (a) Proof Assume that [latex]a[/latex] is the least element of [latex]B[/latex] in [latex]R^{-1}[/latex], then [latex]\forall x\in{B},a\leq{x}[/latex] Now, for [latex]R[/latex] is [latex]\forall x\in{B},x\leq{a}[/latex] Hence [latex]a[/latex] is the greatest element of [latex]B[/latex] in [latex]R[/latex]. The reciprocal proof is similary... (b) Proof For Supremum... Assume that [latex]a[/latex] is the supremum of [latex]B[/latex] in [latex]R^{-1}[/latex], then [latex]\forall x\in{B},x\leq{a}[/latex] For R is [latex]\forall x\in{B},a\leq{x}[/latex] This prove that [latex]a[/latex] is lower bound of [latex]B[/latex] in [latex]R[/latex]. Now only remains to prove that [latex]a[/latex] is the mininum lower bound... How i prove this? Good day Link to comment Share on other sites More sharing options...
DrRocket Posted March 2, 2012 Share Posted March 2, 2012 Now only remains to prove that [latex]a[/latex] is the mininum lower bound... How i prove this? Good day Use the fundamental definition. Link to comment Share on other sites More sharing options...
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