Question

I need some help..

Answer 1

any idea?

Answer 2

let's first deal with one-to-one: let \(S_1,S_2\in D\) such that
\[ \large f(S_1)=f(S_2) \]
we have to prove that \(S_1=S_2\). we have four cases:
\(\mathbf{(1)}\) \(1\in S_1\qquad 1\in S_2\) then
\[ \large f(S_1)=f(S_2) \]
\[ \large S_1\setminus\{1\}=S_2\setminus\{1\} \]
\[ \large S_1=S_2 \]
\(\mathbf{(2)}\) \(1\notin S_1\qquad 1\notin S_2\) then
\[ \large f(S_1)=f(S_2) \]
\[ \large S_1\cup\{1\}=S_2\cup\{1\} \]
\[ \large S_1=S_2 \]

Answer 3

Do we need to prove if \[ S_1 \not=S_2\quad then \quad f(S_1)\not=f(S_2)\] for one to one

Answer 4

actually we can use both of them..

Answer 5

these are definition..
\[ \forall a,b \in A, \;\; f(a)=f(b) \Rightarrow a=b \\

\forall a,b \in A, \;\; a \neq b \Rightarrow f(a) \neq f(b) \]

Answer 6

this is definition for onto
\[
\forall y \in Y, \, \exists x \in X, \;\; f(x)=y \]

Answer 7

thanks. but i know the definitions.

Answer 8

onto is easier: let \(K\in E\) if \(1\notin K\) then
\[ \large K=f(K\cup\{1\}) \]
if \(1\in K\) then
\[ \large K=f(K\setminus\{1\}) \]

Answer 9

thanks a lot..

Answer 10

what are the other two cases..