Question

Let f: S--> T be a function.
C1, C2 \(\subset T\), and D1,D2 \(\subset S\)
Prove:
1/ \(f^{-}(C1\cup C2) = f^{-}(C1) \cup f^{-}(C2)\)
2/\( f(D1\cup D2)=f(D1)\cup f(D2)\)
Please, help

Answer 1

@zepdrix @dan815

Answer 2

Loooooose D:
You've gotten too smart!
I can't keep up anymore lol

Answer 3

I am a big LOSER hehehe need a lot of heeeeeeeeeeelp

Answer 4

Please leave instruction here. I appreciate any help. I 'll be back tomorrow. It's midnight here. hihihihi

Answer 5

These are pre-images correct?

Answer 6

Let \(x\in f^{-1}(C_1\cup C_2)\).
Then \(f(x) \in C_1\cup C_2\).
So \(f(x) \in C_1\) or \(f(x) \in C_2\).
Thus \(x\in f^{-1}(C_1)\cup f^{-1}(C_2)\).
So \(f^{-1}(C_1\cup C_2)\subseteq f^{-1}(C_1)\cup f^{-1}(C_2)\)

Now, let \(x\in f^{-1}(C_1)\cup f^{-1}(C_2) \).
Then \(f(x)^{-1}\in C_1 \) or \(f(x)^{-1}\in C_2 \)
Thus \(f(x) \in C_1\) or \(f(x) \in C_2\) i.e. \(f(x) \in C_1\cup C_2\).
So \(x\in f^{-1}(C_1\cup C_2)\). This shows \(f^{-1}(C_1\cup C_2)\supseteq f^{-1}(C_1)\cup f^{-1}(C_2)\).

So \(f^{-1}(C_1\cup C_2) = f^{-1}(C_1)\cup f^{-1}(C_2)\).

The second one will be similar but easier.

Answer 7

note: Those two \(f(x)^{-1}\) should be \(f^{-1}(x)\)

Answer 8

I guess the first one is actually easier.

For the second one.

Let \(y\in f(D_1\cup D_2)\). Then there exists \(x\in D_1\cup D_2\) such that \(f(x) = y\).
Now \(x\in D_1\cup D_2 \implies x\in D_1\) or \(x\in D_2\) i.e. \(f(x) \in f(D_1)\) or \(f(x) \in D_2\implies f(x) \in f(D_1)\cup f(D_2)\iff y\in f(D_1)\cup f(D)_2)\).
So \(f(D_1\cup D_2) \subseteq f(D_1)\cup f(D_2).\)

Now do something similar to show \(f(D_1\cup D_2) \supseteq f(D_1)\cup f(D_2)\).

Answer 9

Thank you @zz0ck3r . I got you.