Question

f:S-->T
show equivalence of
a) f is one to one
b) for every y in T, the set \(f^{-1}(\{y\})\) contains at most 1 point
c) \(f(D_1\cap D_2)=f(D-1)\cap f(D_2)\) for all subsets \(D_1,D_2\) of S
Please, help

Answer 1

@nerdguy2535 can you help me, please?

Answer 2

Alright, as for a plan of attack, we will show: $$a)\Longrightarrow b)\Longrightarrow c)\Longrightarrow a).$$ This will prove all the statements are equivalent.

Answer 3

Step one: If the function f is one to one, then the set \(f^{-1}\left(\{y\}\right)\) has at most one point for every \(y\in T \).

Sketch of proof: Try a proof by contradiction. Assume f is one to one and there exists a \(y_0\in T\) such that the preimage \(f^{-1}\left(\{y_0\}\right)=\{x\in S\mid f(x)=y_0\}\) has more than one element in it, say \(x_1,x_2 \in f^{-1}\left(\{y\}\right), x_1\ne x_2.\) Then what can you say about \(f(x_1)\) and \(f(x_2)\)? Can this function still be one to one?

Answer 4

Step 2: If the set \(f^{-1}\left(\{y\}\right)\) has at most one point for every \(y\in T\), then \(f\left(D_1\cap D_2\right)=f\left(D_1\right)\cap f\left(D_2\right)\) for all subsets \(D_1,D_2\subseteq S\).

Sketch of proof: Here you want to do a double inclusion; show that \(f\left(D_1\cap D_2\right)\subseteq f\left(D_1\right)\cap f\left(D_2\right)\) and \(f\left(D_1\cap D_2\right)\supseteq f\left(D_1\right)\cap f\left(D_2\right)\).

For the first one, you don't need the fact that f is one to one. In fact, for any function f, the statement: $$f\left(D_1\cap D_2\right)\subseteq f\left(D_1\right)\cap f\left(D_2\right)$$is true. Its the reverse direction that uses the fact that f is one to one.

Let \(y\in f\left(D_1\right)\cap f\left(D_2\right)\). Then \(y\in f(D_1)\) and \(y\in f(D_2 )\), which implies that there exists \(x_1\in D_1 \) and \(x_2\in D_2 \) such that \(f(x_1)=f(x_2)=y\). Since f is one to one, what can we say about \(x_1 \) and \(x_2\)?

Answer 5

Step 3: If \(f\left(D_1\cap D_2\right)=f\left(D_1\right)\cap f\left(D_2\right)\) for all subsets of S, then \(f\) is one to one.

Sketch of proof: To use the hypothesis here, we want to make a clever choice for the subsets \(D_1,D_2\) of S. Assume \(x_1\ne x_2.\) If \(y_1=f(x_1)\) and \(y_2=f(x_2),\) we want to show that \(y_1\ne y_2\).

Take \(D_1=f^{-1}\left(\{y_1\}\right)\) and \(D_2=f^{-1}\left(\{y_2\}\right)\). Try plugging these two sets into our hypothesis, and use the fact that: $$f^{-1}\left(A\right)\cap f^{-1}\left(B\right)=f^{-1}\left(A\cap B\right)$$and: $$f\left(f^{-1}\left(\{y\}\right)\right)=\{y\}$$ for any subsets A and B.

Answer 6

With all three of these things proved, you will have your equivalence.

Answer 7

I'm sorry for being late. The net is so bad here.Thanks for your help. :)