in Set Theory & Algebra edited by
379 views
3 votes
3 votes

Suppose that $f$ is a function from $A$ to $B$.We define the function $S_f$ from $P(A)$ to $P(B)$ by the rule $S_f (X) = f (X)$ for each subset $X$ of $A$. Similarly, we define the function $S_f^{-1}$ from P(B) to P(A) by the rule $S_f^{-1}(Y ) = f^{−1}(Y )$ for each subset $Y$ of $B$.



Q. Suppose that $f$ is a function from the set $A$ to the set $B$. Prove that
a) if $f$ is one-to-one, then $S_f$ is a one-to-one function from $P(A)$ to $P(B)$.
b) if $f$ is onto function, then $S_f$ is an onto function from $P(A)$ to $P(B)$.
c) if $f$ is onto function, then $S_f^{-1}$ is a one-to-one function from $P(B)$ to $P(A)$.
d) if $f$ is one-to-one, then $S_f^{-1}$ is an onto function from $P(B)$ to $P(A)$.
e) if $f$ is a one-to-one correspondence, then $S_f$ is a one-to-one correspondence from $P(A)$ to $P(B)$ and $S_f^{-1}$ is a one-to-one correspondence from $P(B)$ to $P(A)$.



[Hint: Use parts (a)-(d).]

in Set Theory & Algebra edited by
379 views

2 Comments

Nice Hint pooja.. laugh

0
0
If inverse is defined for f() and Sf(), then they must be bijective already. So |A| = |B| and |P(A)|= |P(B)|.

Am i right?
0
0

Please log in or register to answer this question.

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true