# Math Help - Help with Intro to Abstract Math Class...

1. ## Help with Intro to Abstract Math Class...

Let A and B be sets and X and Y be subsets of A. Let f : A --> B be
a function. Prove that f(X) - f(Y) is a subset of f(X - Y).

I'm not very comfortable with sets to begin with, but now that I've got functions thrown in and a really bad textbook, I'm lost. I've only gotten to "Let a be a subset of f(X) - f(Y). Then a is in f(X) and a is not in f(Y)." but now I'm not sure where to go next. Any help at all would be greatly appreciated. Thanks. )

2. Originally Posted by erinthered
Let A and B be sets and X and Y be subsets of A. Let f : A --> B be
a function. Prove that f(X) - f(Y) is a subset of f(X - Y).

I'm not very comfortable with sets to begin with, but now that I've got functions thrown in and a really bad textbook, I'm lost. I've only gotten to "Let a be a subset of f(X) - f(Y). Then a is in f(X) and a is not in f(Y)." but now I'm not sure where to go next. Any help at all would be greatly appreciated. Thanks. )
I do not know what you mean by $f(x)$ and $f(x-y)$. Can you explain yourself?

Let me try it like this:
1) $f:A\to B$ (function)
2) $X\subseteq A$ (subset)
3)Then, $f[X]=\{f(x)|x\in X\}$
4)Thus, $f[X]\subseteq B$
Because, since $f(x),x\in A$ we have $f(x)\in B$ because this function maps an element from A into B. But, $x\in X$ implies $x\in A$ thus, for all $x\in X$ we have, $f(x)\in B$. Thus, for all $e\in f[X]$ we have $e\in B$ thus, $f[X]\subseteq B$
But I do not know if you were asking this?

3. f(X) is the image of X under f according to my textbook, same with f(Y). I'm thinking that f(X-Y) is supposed to be the image of the complement of Y in X and I have to prove that it's a subset of the complement of the image of Y in the image of X.

I was trying to pick a random arbitrary element in the left hand side, then prove that it's also in the right hand side. I'm just not sure how to go about doing that.

4. My proof so far (after corrected a stupid mistake.....) is:

Let a be an element of f(X) - f(Y). Then a is in f(X) and is not in f(Y). Since a is in f(X), then a = f(x) for some x in X. Since x is in X, x is in X-Y. Hence a is an element of f(X-Y) and f(X) - f(Y) is a subset of f(X-Y).

I think I'm skipping something or doing something wrong, because it just looks simpler than it should. Bleh. I don't know.

5. Originally Posted by erinthered
f(X) is the image of X under f according to my textbook, same with f(Y). I'm thinking that f(X-Y) is supposed to be the image of the complement of Y in X and I have to prove that it's a subset of the complement of the image of Y in the image of X.

I was trying to pick a random arbitrary element in the left hand side, then prove that it's also in the right hand side. I'm just not sure how to go about doing that.
Okay I get you know, I was close though.

Proofs on set theory I like to write out in works, or in easily understandable sets.
You have $f:A\to B$
$X,Y\subseteq A$
Thus, $X-Y=\{x\in X, x\not \in Y\}$
Thus, $F[X-Y]=\{f(x)|x\in X,x\not \in Y\}$
Thus, $F[X]=\{f(x)|x\in X\}$
Thus, $F[Y]=\{f(x)|x\in Y\}$
Thus, $F[X]-F[Y]=\{f(x)|f(x)\in F[X],f(x)\not \in F[Y]\}$
You need to show that,
$\{f(x)|f(x)\in F[X],f(x)\not \in F[Y]\}\subseteq \{f(x)|x\in X,x\not \in Y\}$

Which is true, you see why?
(I just cannot complete the proof. SO tired now- got a bussiness law test soon. Got to go bye.)

6. yay! I see it clearly now instead of the vague fuzz that I was seeing. Thanks. Luck on the Business Law test.

7. Originally Posted by erinthered
yay! I see it clearly now instead of the vague fuzz that I was seeing. Thanks. Luck on the Business Law test.
1) $F[X]-F[Y]$ empty.
2) $F[X]-F[Y]$ non-empty.
If 2 you need to show if $x\in F[X]-F[Y]$ then $x\in F[X-Y]$ then by definition of subset we have $F[X]-F[Y]\subseteq F[X-Y]$
Finally, always know the difference between $\subset$ and $\subseteq$. They mean different things.