COT5310hw5key - COT 5310 Homework 5 Key Fall 2007 1 Let INF...

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COT 5310 Homework 5 Key Fall 2007 1. Let INF = { f : dom( f ) is infinite } and NE = { f : there is a y such that f ( y ) converges } . Show that NE m INF. Present the mapping and then explain why it works as desired. To do this define a total recursive function g such that index f is in NE iff g ( f ) is in INF. Be sure to address both cases ( f in and f not in). We need a mapping g such that f NE g ( f ) INF. So we don’t write g ( f )( x ) we’ll use the notation g f = g ( f ). Define g f to be g f ( ( x, t ) ) = μ z [STP( x, f, t )] . So g f ( ( x, t ) ) returns 0 if STP( x, f, t ) otherwise it diverges. If f NE there exists a y such that f ( y ) converges, meaning there exists a ( y, t ) such that STP( y, f, t ) is true. Then since STP( y, f, t ) STP( y, f, t + k ) for all k 0, g f INF. If f / NE then there does not exist a y such that f ( y ) converges, meaning that dom( f ) = . This means that for all ( y, t ) , STP( y, f, t ) is false. Since the domain of g f in this case has size 0, g f / INF.
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