Question

tthese two are related

(5 points.) Since A) is a set, we can inquire, for each element x of S, whether x is an element of R) We define the set N to be the set of all elements of S for which the statement "x is an element of Ax)" is NOT true: N = {x: x is an element of S and x is NOT an element of Ax)}. Explain why N is an element of P(S). This set N, defined for a given function F as described above, will be referred to again in Problems 3 and 4 on this page. We now have a set N [see Problem 21 which is a subset of S and also an element of the codomain of F: S- P(S). Let us think about what happens if there exists an element c in S such that Rd = N. In (a) and (b) below, we investigate the consequences of this supposition. a. Is it possible to have c be an element of N? Explain (of course). b s it possible to have c NOT be an element of N? Explain (of course). Hint: In BOTH cases, think carefully about the consequences under the initial assumption that the correct answer is "yes c. What conclusion do you draw from the results of (a) and (b)? (Does there exist an element c in S such that A = N, as we assumed above?? Is N in the range of Note the question is, what can you conclude from (a) and (b).. with this entire Project, your answers should be based on what is in the Project, not outside ideas.

Answer 1

F S P(s) is function lement of Sand N= {xE S: is an lemert fF)} is not xeS xF«)} C S PS) N E F CE St F=N NowBuppo FC) N 4 Suppo CEN N C E xE S: F(x)}ine ceFC) CEN is not tue N cs Scanned with CamScande

Now Auppos cf N b) e c xeS: x Fx) r.e But F(C) N CE N not 1ue is e CE N The conclusions bfCa)b) kad to parader called Russel's paradox Enelement annst belong ompument Aimutaneeush Hence , the assumption is not tue Jhese is FC)= N a t a set and its lmant ce S Buch that QU Note his rores that foA any det S, there is mot surjective undion F s-s Ps) thereis Since CS witpreimage set Ne PS) ohich alunys a canned CamScanner