1. ## Functions Sets

Suppose that

f : A->B is onto and g : A->B is 1-to-1. Let S be a
subset of A. Consider the sets g^-1[f[S]] and f^-1[g[S]]. If they're not the
same size, which one must be larger?

Need alittle help on this problem.

2. ## Re: Functions Sets

Originally Posted by chrisgg1212
If they're not the same size, which one must be larger?
Are both of them finite by assumption? If not, then larger in what sense? It seems to me that it may happen that neither of these sets is contained in the other one.

3. ## Re: Functions Sets

I don't know what you mean.

That is how the question was handed to me for practice...I don't know what to do.

4. ## Re: Functions Sets

practice for what class? Do you not know what "finite sets" are?

5. ## Re: Functions Sets

yes finite # of elements.

6. ## Re: Functions Sets

Originally Posted by chrisgg1212
I don't know what you mean.
That is how the question was handed to me for practice...I don't know what to do.

If the sets are finite then $\displaystyle \|A\|=\|B\|$, that is they have the same cardinality.

If the sets are infinite the look for examples.

7. ## Re: Functions Sets

Hi,
Here's the answer to your question. From your posting, I'm not sure you can understand anything except the example. So I think it was probably an unreasonable problem assignment.

8. ## Re: Functions Sets

Intuitively, there are the following four principles.

(1) The image of a set with respect to any function is no larger than the original set because multiple elements can be mapped into one.

(2) The image of a set with respect to a 1-1 function has the same size as the original set because each element is mapped into exactly one element.

(3) The preimage of a set with respect to an onto function is no smaller then the original set because every element has at least one preimage and possible more.

(4) The preimage of a set with respect to a 1-1 function is no larger than the original set because not every element may have a preimage (unless the function is also onto), and if there is a preimage, it is unique.

Using these principles, we have $\displaystyle |g^{-1}[f[S]]| \stackrel{(4)}{\le} |f[S]| \stackrel{(1)}{\le} |S| \stackrel{(2)}{=\vphantom{\le}} |g[S]| \stackrel{(3)}{\le} |f^{-1}[g[S]]|$.