Question

5. (10 pts) Let N be the natural numbers with the usual ordering and let A = NU{2} be ordered as 1, 2, 3, .. 9 2 (a) Is there a bijection g:N + A? (Why.why not?) (b) Is there an order isomorphism f:N → A? (Why/why not?)

Answer 1

since g is strictly increasing, so for X1<=X2, g(X1)<=g(X2)

n2-1 고. 5. Let iN be the natural numbers with the usual ordering and let Az NV44} be ordered as 1,2,3,...,Ż a) Let g:N → A be defined as g (n) = { for n=1 =n- for n>,2 for Ni, NZEN if g (n.) = g(^2) = 1/2 then ni=92=1 if g (ni) = g(^2)+1 then 9,-) => n = n2 50 g(x) = g(y) = x=g ¥xy EIN o. 9 is injective, For ļ EA , 901) = Where ne IN for nea, n+1 = EA , 3XEIN sit g(x) = y ing is burjective :. There is a bijedion froom g: INA. b) Two totally ordered sets (A, 4) and (Bs)are order inomor phie iff there is a bijection f from A to such that for all as, az EA, a, s az iff fa) f(az) If we consider f:N → A same as the function 9:INA defined above ni = n2 iff g (n.) = g(22) & n., n EIN :. There is an order isomorphism FPANA. :. For any B