# Thread: Proving Subset

1. ## Proving Subset

Prove: If $\displaystyle f: A \rightarrow B$ and $\displaystyle g: B \rightarrow C$ are functions and $\displaystyle D \subset A,$ then $\displaystyle (g \circ f) (D) = g(f(D))$.

2. Originally Posted by txsoutherngirl84
Prove: If $\displaystyle f: A \rightarrow B$ and $\displaystyle g: B \rightarrow C$ are functions and $\displaystyle D \subset A,$ then $\displaystyle (g \circ f) (D) = g(f(D))$.
.....isn't that the definition?

3. Actually it is not the definition.
$\displaystyle g \circ f(D)$ is the image of $\displaystyle D$ under the mapping $\displaystyle g \circ f:A\to C$
Whereas, $\displaystyle g(f(D))$ is the image of $\displaystyle f(D)$ under the mapping $\displaystyle g:B\to C$.

If $\displaystyle p\in g \circ f(D)$ then by definition $\displaystyle \left( {\exists d \in D} \right)\left[ {g \circ f(d) = p} \right]$.
Also $\displaystyle \left( {\exists b \in B} \right)\left[ {f(d) = b \wedge g(b) = p} \right]$,which means $\displaystyle p\in g(f(D))$

4. Originally Posted by Plato
Actually it is not the definition.
$\displaystyle g \circ f(D)$ is the image of $\displaystyle D$ under the mapping $\displaystyle g \circ f:A\to C$
Whereas, $\displaystyle g(f(D))$ is the image of $\displaystyle f(D)$ under the mapping $\displaystyle g:B\to C$.

If $\displaystyle p\in g \circ f(D)$ then by definition $\displaystyle \left( {\exists d \in D} \right)\left[ {g \circ f(d) = p} \right]$.
Also $\displaystyle \left( {\exists b \in B} \right)\left[ {f(d) = b \wedge g(b) = p} \right]$,which means $\displaystyle p\in g(f(D))$
Excuse me, let me rephrase that. "Isn't that almost everyone's definition? Not because it is the formal definition but it blindingly obvious?"

5. Originally Posted by Drexel28
Excuse me, let me rephrase that. "Isn't that almost everyone's definition? Not because it is the formal definition but it blindingly obvious?"
But that is exactly the point. This problem deals with a formal definition.
Now I know that this is a personal peeve of mine.
I give Charles Pinter full credit for making me aware of this problem.
It is a notational problem.
Suppose that $\displaystyle f:A\to B$ and $\displaystyle C\subset A$ what sense does it make to write $\displaystyle f(C)?$
I agree with him, it makes no sense. But I disagree with his notation.
He uses $\displaystyle \overline f (C)$ for $\displaystyle f(C)$.
And if $\displaystyle D\subset B$ then $\displaystyle \overline{\overline f} (D)$ is used for $\displaystyle f^{-1}(D)$.
I suggested the more intuitive notations $\displaystyle \mathop f\limits^ \to (C)\,\& \,\mathop f\limits^ \leftarrow (D)$ but never got any real takers.

6. Originally Posted by Plato
But that is exactly the point. This problem deals with a formal definition.
Now I know that this is a personal peeve of mine.
I give Charles Pinter full credit for making me aware of this problem.
It is a notational problem.
Suppose that $\displaystyle f:A\to B$ and $\displaystyle C\subset A$ what sense does it make to write $\displaystyle f(C)?$
I agree with him, it makes no sense. But I disagree with his notation.
He uses $\displaystyle \overline f (C)$ for $\displaystyle f(C)$.
And if $\displaystyle D\subset B$ then $\displaystyle \overline{\overline f} (D)$ is used for $\displaystyle f^{-1}(D)$.
I suggested the more intuitive notations $\displaystyle \mathop f\limits^ \to (C)\,\& \,\mathop f\limits^ \leftarrow (D)$ but never got any real takers.
I actually like that notation it makes more sense. I guess the real idea is that given a function $\displaystyle f:A\to B$ you have automatically induced a natural function $\displaystyle \vec{f}:\mathcal{P}(A)\to\mathcal{P}(B):E\mapsto f(E)$ and then once you have that nasty notational buisness out of the way you can say $\displaystyle \vec{f}(E)$ and not have people wonder why a function which maps elements of a set is now mapping full subsets of that set.