1. ## Set Proof check

I need to prove each side is a subset of the other.

Let F be a function. Show that $F(A$ U $B)$ = $F(A)$ U $F(B)$

Let (x,y) be $\in$ $F(A$ U $B)$. Then there exists a point F(X) and a point F(y) such that x or y $\in$ $F(A)$ or $F(B)$. Since (x,y) $\in$ $F(A$ U $B)$ it follows (x,y) $\in$ $F(A)$ U $F(B)$.

Am i on the right track?
I did the other half but I did not want to waste time typing it if this was wrong, since I used a similar method.

2. ## Set Proof

Hello p00ndawg
Originally Posted by p00ndawg
I need to prove each side is a subset of the other.

Let F be a function. Show that $F(A$ U $B)$ = $F(A)$ U $F(B)$

Let (x,y) be $\in$ $F(A$ U $B)$. Then there exists a point F(X) and a point F(y) such that x or y $\in$ $F(A)$ or $F(B)$. Since (x,y) $\in$ $F(A$ U $B)$ it follows (x,y) $\in$ $F(A)$ U $F(B)$.

Am i on the right track?
I did the other half but I did not want to waste time typing it if this was wrong, since I used a similar method.
You're kind of on the right track, but you're not using the notation correctly. So let me start you off.

As you say, we need to prove that each set is a subset of the other, in order to prove they're equal. So let's prove first that $F(A \cup B) \subseteq (F(A) \cup F(B))$. We do this by choosing an element in the first set, and showing that it's also in the second.

So let's suppose that $y \in F(A \cup B)$. That means that there's an $x \in (A \cup B)$ for which $y = F(x)$.

Now $[x \in (A \cup B)] \Rightarrow [(x \in A) \lor (x\in B)]$

But $[x \in A]\Rightarrow [y \in F(A)]$ and $[x \in B]\Rightarrow [y \in F(B)]$

So $[x \in (A \cup B)]\Rightarrow [(y \in F(A)) \lor (y \in F(B))]$

$\Rightarrow y \in (F(A) \cup F(B))$

So we've just proved that $[y \in F(A \cup B)] \Rightarrow [y \in (F(A) \cup F(B))]$. In other words, that $F(A \cup B) \subseteq (F(A) \cup F(B))$

OK? Now you've got to prove it the other way round. Start with $y \in (F(A) \cup F(B))$ and $y = F(x)$ ...

3. ok lets see if this is right...

Suppose $Y \in F(A) \cup F(B)$and $y = F(x).$
Since $y \in F(A) \cup F(B)$, $Y \in F(A)$ or $Y \in F(B)$.
If $Y \in F(A)$, then $x \in A$, and if $y \in F(B)$then $x \in B$.
Because $y \in F(A) \cup F(B)$, it follows that $x \in A$ or $x \in B$.

Thus, $F(A) \cup F(B) \subseteq F(A \cup B)$.

4. Originally Posted by p00ndawg
Suppose $Y \in F(A) \cup F(B)$and $y = F(x).$
Since $y \in F(A) \cup F(B)$, $Y \in F(A)$ or $Y \in F(B)$.
If $Y \in F(A)$, then $x \in A$, and if $y \in F(B)$then $x \in B$.
Because $y \in F(A) \cup F(B)$, it follows that $x \in A$ or $x \in B$.

Thus, $F(A) \cup F(B) \subseteq F(A \cup B)$.
I think that you have a great deal of work to do on understanding the notation involved in this question.
There is no way to prove something if you cannot use the correct notation.

5. Originally Posted by Plato
I think that you have a great deal of work to do on understanding the notation involved in this question.
There is no way to prove something if you cannot use the correct notation.
what do you mean by notation?

6. Hello p00ndawg
Originally Posted by p00ndawg
ok lets see if this is right...

Suppose $Y \in F(A) \cup F(B)$and $y = F(x).$
Since $y \in F(A) \cup F(B)$, $Y \in F(A)$ or $Y \in F(B)$.
If $Y \in F(A)$, then $x \in A$, and if $y \in F(B)$then $x \in B$.
Because $y \in F(A) \cup F(B)$, it follows that $x \in A$ or $x \in B$.

Thus, $F(A) \cup F(B) \subseteq F(A \cup B)$.
This is essentially correct, although not very elegant.

Originally Posted by Plato
I think that you have a great deal of work to do on understanding the notation involved in this question.
There is no way to prove something if you cannot use the correct notation.
Perhaps Plato would give us his version of the proof?

7. Suppose $Y \in F(A) \cup F(B) \Rightarrow$ $x \in A$or $x \in B$ for $y = F(x).$
If $y \in F(A) \cup F(B)$, then $x \in A$ or $y \in F(A)$, if $x \in B$ then $y \in F(B)$.
Because $x \in A$ or $x \in B$, then $y \in F(A)$or $y \in F(B)$.
It follows that $F(A) \cup F(B) \subseteq F(A \cup B)$.