1. ## General and subsets

Have to determine whether each one is true or false.

For #6. I used a similar proof for F(AUB). Had this:
suppose y∈(A⋂B)={F(x)|x∈A⋂B}
Then y=F(x) for some x∈A⋂B
y=F(x) where x∈A or x∈B
But x∈A---> y=F(x)∈F(A)
and x∈B--> y=F(x)∈F(B)
y∈F(A) and y∈F(B)
y∈F(A)⋂F(B)

is this correct or is this different way for ⋂?

2. Originally Posted by hellfire127
Have to determine whether each one is true or false.

For #6. I used a similar proof for F(AUB). Had this:
suppose y∈(A⋂B)={F(x)|x∈A⋂B}
Then y=F(x) for some x∈A⋂B
y=F(x) where x∈A or x∈B
But x∈A---> y=F(x)∈F(A)
and x∈B--> y=F(x)∈F(B)
y∈F(A) and y∈F(B)
y∈F(A)⋂F(B)

is this correct or is this different way for ⋂?
Your proof is incomplete for number 6. You have proven that $F\left(A\cap B\right)\subseteq A\cap B$. But, the reverse inclusion is not true. Consider a non-injective function (such as $f-1,1)\to(-1,1):x\mapsto x^2" alt="f-1,1)\to(-1,1):x\mapsto x^2" />).

For the second one, are you having trouble? If so, where at?

3. yea just found out that number 6 is false using A as {1,} and B as {1,3} for example.

yes, i am having trouble understading the inverse for C and D on number 7.

4. Originally Posted by hellfire127
yea just found out that number 6 is false using A as {1,} and B as {1,3} for example.

yes, i am having trouble understading the inverse for C and D on number 7.
What do you mean the 'inverse'?

5. Originally Posted by Drexel28
What do you mean the 'inverse'?
the power to -1
F^-1

6. Originally Posted by hellfire127
the power to -1
F^-1
It's not a literal inverse, it's the inverse image. Namely, $F^{-1}\left(A\right)=\left\{x:F(x)\in A\right\}$

7. yea meant inverse image. i really need to go back and read the section over on this as I don't even know how to start most of these.

8. Originally Posted by hellfire127
yea meant inverse image. i really need to go back and read the section over on this as I don't even know how to start most of these.
Well, what if I get you started?

If $x\in F^{-1}(U\cap V\right)$ then $F(x)\in U\cap V$ and so $F(x)\in U\text{ and }F(x)\in V$ and so $x\in F^{-1}(U)\text{ and }x\in F^{-1}(V)$ and so $x\in F^{-1}(U)\cap F^{-1}(V)$.

9. Originally Posted by Drexel28
Well, what if I get you started?

If $x\in F^{-1}(U\cap V\right)$ then $F(x)\in U\cap V$ and so $F(x)\in U\text{ and }F(x)\in V$ and so $x\in F^{-1}(U)\text{ and }x\in F^{-1}(V)$ and so $x\in F^{-1}(U)\cap F^{-1}(V)$.
so whenever there is an inverse image for $x\in F^{-1}(A\cap B\right)$ then $F(x)\in A\cap B$ ? like the A and B can be any sets correct?

10. Originally Posted by hellfire127
so whenever there is an inverse image for $x\in F^{-1}(U\cap V\right)$ then $F(x)\in A\cap B$ ? like the A and B can be any sets correct?
Yes, because this is the definition.

11. ok this seems a lot harder compared to mathematical induction earlier.

my book has a smiliar problem to number 6 but i don't understand its prove. do you have a better explaination for number 6? i just can't seem to see how it works out. i understand your solution to number 7 more.

12. Originally Posted by hellfire127
ok this seems a lot harder compared to mathematical induction earlier.

my book has a smiliar problem to number 6 but i don't understand its prove. do you have a better explaination for number 6? i just can't seem to see how it works out. i understand your solution to number 7 more.
What do you mean explanation? You proved that the inclusion goes one way, but not the other. What more do you seek?

13. like something you did smilar to number 7 for number 6.

14. Originally Posted by hellfire127
like something you did smilar to number 7 for number 6.