1 votes 1 votes Show that if $S$ is a set, then there does not exist an onto function $f$ from $S$ to $P(S),$ the power set of $S$. Conclude that $\mid S\mid < \mid P(S)\mid .$ This result is known as Cantor’s theorem. [Hint: Suppose such a function $f$ existed. Let $T = \{s \in S \mid s \notin f (s)\}$ and show that no element $s$ can exist for which $f (s) = T.]$ Set Theory & Algebra kenneth-rosen discrete-mathematics set-theory&algebra descriptive + – admin asked Apr 21, 2020 • edited Apr 21, 2020 by Lakshman Bhaiya admin 657 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes S to P(S) no onto function is possible. because if |S|=n then |P(S)|=2^n. and these two can never be equal. always |S| <|P(S)|. but onto function condition is if there is an onto function from A to B then |A|>=|B|. but here the case is just reverse. hence S to P(S) no onto function is possible. debasree88 answered Apr 22, 2020 debasree88 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes see in simple words lets take an example s{1,2,} p(s)={phai,{1},{2},{1,2}} now in onto BASE CONDITION IS IF THERE EXIST A ONTO FXN BETWEEN A TO B THEN |A|>=|B| but s to p(s) condn gets fails so ONTO (SURJECTION) NOT POSSIBLE HERE lovegate answered Mar 9, 2021 lovegate comment Share Follow See all 0 reply Please log in or register to add a comment.