FallArk said:
Find a function g from {0,1} to B\A such that f^-1(g(x)) = x +2 for x∈{0,1}. Present it in the 2-row form.
A = {{1},2,3} and B = {∅,1,{2},3}
I know that B\A = {∅,1,{2}} and f is a bijection from A to B\A
how do I find such function g? It obviously can't be bijection, how do I match one value to another?
Hey FallArk! (Smile)
If I understand correctly, we're talking about something like this:
\begin{tikzpicture}
%preamble \usetikzlibrary{shapes.geometric,fit}
\foreach \x in {0,1}{
\node[fill,circle,inner sep=2pt, label=left:\x] (d\x) at (0,\x) {};
}
\node[fit=(d0) (d1),ellipse,draw,minimum width=1.5cm] (D) {};
\foreach \x[count=\xi] in {0,1,2}{
\node[fill,circle,inner sep=2pt] (b\xi) at (4,\x) {};
}
\node[fit=(b1) (b2) (b3),ellipse,draw,minimum width=1.5cm] (BmA) {};
\node[above] at (BmA.north) {$B \setminus A$};
\foreach \x/\p[count=\xi] in {1/\{1\},2,3}{
\node[fill,circle,inner sep=2pt, label=right:\p] (a\xi) at (8,\x) {};
}
\node[fit=(a1) (a2) (a3),ellipse,draw,minimum width=1.5cm] (A) {};
\node[above] at (A.north) {$A$};
\draw[-latex] (d0) -- node[below] {$g(0)$} (b1);
\draw[-latex] (a2) -- node[above,xshift=8,yshift=12] {$f(2+0)$} (b1);
\draw[-latex] (d1) -- node[above] {$g(1)$} (b3);
\draw[-latex] (a3) -- node[above] {$f(2+1)$} (b3);
\draw[-latex] (a1) -- node[below right,xshift=10] {$f(1)$} (b2);
\end{tikzpicture}
However, we don't seem to have any information about which elements in $B \setminus A$ we should link to, do we? (Wondering)