# Thread: How to prove a statement of maps and sets?

1. ## How to prove a statement of maps and sets?

Hi everyone,
I need some help on an exercise because i didn't really understand how to prove a statement like this :
let E and F be two sets and let $f$ be a map from E to F. Let A and A' be two subsets of E. Let B and B' be two subsets of F.

$f$(AUA')=( $f$(A)U $f$(A'))

Thanks a lot

2. ## Re: How to prove a statement of maps and sets?

the usual way to prove 2 sets are equal is to show that each one contains the other.

so in order to prove that A = B, you show that any a in A is also in B, and that any b in B is also in A.

suppose x is in f(A U A'). this means that x = f(u), for some u that is in A U A'.

u being in A U A', means that either u is in A, OR u is in A' (or maybe both).

to show that x is in f(A) U f(A'), you need to show that either x is in f(A) (that is x = f(a) for some a in A), OR x is in f(A') (that is, x = f(a') for some a' in A').

i recommend you do two cases:

1) u is in A
2) u is in A'

and show that either way...f(u) is in one of f(A) or f(A'). that will be half the proof, then you go "the other way" and start with y in f(A) U f(A').

3. ## Re: How to prove a statement of maps and sets?

Originally Posted by dekl
Hi everyone,
I need some help on an exercise because i didn't really understand how to prove a statement like this :
let E and F be two sets and let $f$ be a map from E to F. Let A and A' be two subsets of E.
$f$(AUA')=( $f$(A)U $f$(A'))
I will do half of the above.
$\begin{array}{*{20}c} {b \in f(A \cup A')} \\ {\left( {\exists a \in (A \cup A')} \right)\left[ {f(a) = b} \right]} \\ {\left[ {a \in A \wedge f(a) = b} \right] \vee \left[ {a \in A' \wedge f(a) = b} \right]} \\ {\left[ {b \in f\left( A \right)} \right] \vee \left[ {b \in f\left( {A'} \right)} \right]} \\ {b \in \left( {f(A) \cup f(A')} \right)} \\\end{array}$

Please, you show us the other half.

4. ## Re: How to prove a statement of maps and sets?

Thank you very much, I am working on it.

5. ## Re: How to prove a statement of maps and sets?

Originally Posted by Plato
I will do half of the above.
$\begin{array}{*{20}c} {b \in f(A \cup A')} \\ {\left( {\exists a \in (A \cup A')} \right)\left[ {f(a) = b} \right]} \\ {\left[ {a \in A \wedge f(a) = b} \right] \vee \left[ {a \in A' \wedge f(a) = b} \right]} \\ {\left[ {b \in f\left( A \right)} \right] \vee \left[ {b \in f\left( {A'} \right)} \right]} \\ {b \in \left( {f(A) \cup f(A')} \right)} \\\end{array}$

Please, you show us the other half.
Sorry but I don't understand why wouldn't it be done like this...

6. ## Re: How to prove a statement of maps and sets?

b is in (f(A) U f(A'))
then there is an a in A or in A' such that f(a)=b
then a is in A U B
so b is in f(A U A')

is that correct?

sorry i don't know how to make thi signes...

7. ## Re: How to prove a statement of maps and sets?

Originally Posted by dekl
b is in (f(A) U f(A'))
then there is an a in A or in A' such that f(a)=b
then a is in A U B so b is in f(A U A')
When proving statement about image is more difficult than is about preimages.
The reason is the use of the existential operator.
In this case one would have to say:
$\left( {\exists a \in A} \right)\left[ {f(a) = b} \right] \vee \left( {\exists a' \in A'} \right)\left[ {f(a') = b} \right]$.
Now note that $\left( {a \in \left( {A \cup A'} \right)} \right) \vee \left( {a' \in \left( {A \cup A'} \right)} \right)$

8. ## Re: How to prove a statement of maps and sets?

thank you verr much.

now can you tell me if I am wrong for the next one?
f-1(B U B') = (F-1(B) U f-1(B'))
this means that f(x) B U B' = f(x)B ou f(x) B'
let take z in B U B' such that f(x)=z
B U B' means that z is in B or z is in B'
then we have f(x)B ou f(x) B'
so f-1(B U B') = (F-1(B) U f-1(B'))

9. ## Re: How to prove a statement of maps and sets?

Originally Posted by dekl
now can you tell me if I am wrong for the next one?
f-1(B U B') = (F-1(B) U f-1(B'))
this means that f(x) B U B' = f(x)B ou f(x) B'
let take z in B U B' such that f(x)=z
B U B' means that z is in B or z is in B'
then we have f(x)B ou f(x) B'
so f-1(B U B') = (F-1(B) U f-1(B'))
Yes, your basic idea is correct.

You can learn LaTeX code.
[TEX]f^{-1}(B\cup B')=f^{-1}(B)\cup f^{-1}(B')[/TEX] gives
$f^{-1}(B\cup B')=f^{-1}(B)\cup f^{-1}(B')$.