# Help with Intro to Abstract Math Class...

• Mar 19th 2006, 07:09 PM
erinthered
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. :o)
• Mar 19th 2006, 07:58 PM
ThePerfectHacker
Quote:

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. :o)

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?
• Mar 19th 2006, 08:22 PM
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.
• Mar 19th 2006, 08:34 PM
erinthered
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.
• Mar 19th 2006, 08:36 PM
ThePerfectHacker
Quote:

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.)
• Mar 19th 2006, 08:56 PM
erinthered
yay! I see it clearly now instead of the vague fuzz that I was seeing. Thanks. Luck on the Business Law test.
• Mar 20th 2006, 03:52 PM
ThePerfectHacker
Quote:

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.