Find g: Solving a Set Theory Problem

[FallArk]

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?

 

[FallArk]

I figured it out, i just have to trace it using the function given to find for x=0 and 1 what is the corresponding values using the two row representation of f

 

[FallArk]

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?

Given A X (A∩B) = { ({1},3), (2,3), (3,3) }, and P(A)\B={ {1}, 2, {{1},2}, {{1},3}, {2,3}, {{1},2,3} }
does there exist a surjection from A X (A∩B) to P(A)\B?

 

[I like Serena]

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)

 

[FallArk]

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)

I actually figured it out myself. But then I run into a new problem. Is it possible for a Cartesian product to be mapped into another set? (the third post is the problem)

 

[I like Serena]

I actually figured it out myself. But then I run into a new problem. Is it possible for a Cartesian product to be mapped into another set? (the third post is the problem)

Okay.
For the record, it seems that $P(A)\setminus B$ contains elements that are integers, but $P(A)$ is supposed to only contain sets doesn't it? (Wondering)
I think it should contain {{1}} instead of {1}.
And {2}, which would then be removed by subtracting $B$.

And sure, we can map a cartesian product to another set.
I don't think the example will be surjective though, since the cartesian product contains 3 elements, and the co-domain contains more than 3 elements.

 

[FallArk]

Okay.
For the record, it seems that $P(A)\setminus B$ contains elements that are integers, but $P(A)$ is supposed to only contain sets doesn't it? (Wondering)
I think it should contain {{1}} instead of {1}.
And {2}, which would then be removed by subtracting $B$.

And sure, we can map a cartesian product to another set.
I don't think the example will be surjective though, since the cartesian product contains 3 elements, and the co-domain contains more than 3 elements.

Ugh, this whole power set thing is so confusing... Surjective means something like for one y there could be more than one x right? So since there are more than 3 y-values, and only 3 x-values, it cannot be surjective.

 

[I like Serena]

Ugh, this whole power set thing is so confusing... Surjective means something like for one y there could be more than one x right? So since there are more than 3 y-values, and only 3 x-values, it cannot be surjective.

Yep.
And more precisely, surjective means that for every y there has to be at least one x.