## anonymous 3 years ago I need some help..

1. anonymous

2. anonymous

any idea?

3. anonymous

4. helder_edwin

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 $\large f(S_1)=f(S_2)$ $\large S_1\cup\{1\}=S_2\cup\{1\}$ $\large S_1=S_2$

5. anonymous

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

6. anonymous

actually we can use both of them..

7. anonymous

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

8. anonymous

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

9. helder_edwin

thanks. but i know the definitions.

10. helder_edwin

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

11. anonymous

thanks a lot..

12. anonymous

what are the other two cases..