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helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
let's first deal with onetoone: 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 \[ \large f(S_1)=f(S_2) \] \[ \large S_1\cup\{1\}=S_2\cup\{1\} \] \[ \large S_1=S_2 \]
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
actually we can use both of them..
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
these are definition.. \[ \forall a,b \in A, \;\; f(a)=f(b) \Rightarrow a=b \\ \forall a,b \in A, \;\; a \neq b \Rightarrow f(a) \neq f(b) \]
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
this is definition for onto \[ \forall y \in Y, \, \exists x \in X, \;\; f(x)=y \]
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
thanks. but i know the definitions.
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
onto is easier: let \(K\in E\) if \(1\notin K\) then \[ \large K=f(K\cup\{1\}) \] if \(1\in K\) then \[ \large K=f(K\setminus\{1\}) \]
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
thanks a lot..
 one year ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
what are the other two cases..
 one year ago
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