Question

9 / 12 + 75% Suppose R is a reflexive and symmetric relation on a finite set A. Define a relation Son A by declaring 1Sy if and only if for some n e N there are elements 11, 12,...,In € A satisfying 1R:11, 11 Rr2, Rts, R14, ..., In-1 R1, and Ry. (i) Prove that S is an equivalence relation on A. (6)

Answer 1

asy it Suppore R is a Reflexive and dynnetric Reln on a kinite Set A. had er deline detine a relation s on A A by for some new shone eurt elements ANEA defish is buying aRx, x R xa xq Rx X-1R Xn and an Ry. prove that I is an eq. rel on A. We know that a rel" is Ree it; a> olis Reblexive 5) Symmetic > Transitive >> we have to prove .

x = 1,2---n. Reflexive: het A={x, da ... n} clearly asxi Since for neicol There evol aclement XiEA X; RX; . dalistying So Hi S Dei V 1,2-n e Hi EA. So I is Reflexive 6 Symmetric had a sy So for Some new cheere uist elements x, 4p, ... REA Classbying x Ra, &, Rai xn-, Rao and a Since R is Symmetric So n. Ru, xQ R x,, XnRXna, and y R an. x, Ry. hat do = sli, an-1= x, de 2 = x; - X, = 3

so y Rxi, x' Rx, xj Rx - and xn RX So clearly y SX for Some ned Share onists xn' do lesbying g Rai, & Rh xus RXn and on Ra wer. So y Esi So & x, y E dif xs y y S x is Synmeloc. x Szi o there onists dince & Transitive hedux, y, z EA sit & Sy & y sz We have to show that s is fransitive ie to prove for Some x sy noor da, EA Salistying xra, x, RY Xanga Rih, and an Ry Since yaz do for Some ng Ed Shene enish Ing EA Sadistying y Ry, I. RG The Ryou and in az XX -- Y, Y2

neol, noch and led So that en hel na nangten So kom dome near there exisl elements 3, 2, 3 - - In, Y, Y., Y, Yong E A Salisbying Ho, Ry, Y RY, , Y, Ry - - x Ru, H, RA Rz So clearly & sz. n. Y, Z CA st x 59 & 9 Bz Since & => x se on D So d is deansidine. Cinco S is Reflenire , Symm. and transitive So By defn: S is equivalence Rel on A.

on R bo a Shen R is Reblenire ✓ Summary: het Deln Sed A; A- tinide. any Deblexive is het A={a, anna an} of di Rai Symmetric - A Rein R on A is doid to be Symmefreic it for all (ay) E B => (y,x) ER. Transitive: A Reln a on A is hard to be Transifive its (x,y) ER and (Y,Z) ER = (x2) ER 6 A Reen is Deti, Symm, drans. Then the hell is equivalence Del 2, 4, Z CA XY EA st for all