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.

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- April 22nd 2013, 11:07 AMchrisgg1212Functions 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. - April 22nd 2013, 11:55 AMemakarovRe: Functions Sets
- April 22nd 2013, 03:50 PMchrisgg1212Re: 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. - April 22nd 2013, 04:20 PMHallsofIvyRe: Functions Sets
practice for what class? Do you not know what "finite sets" are?

- April 22nd 2013, 04:45 PMchrisgg1212Re: Functions Sets
yes finite # of elements.

- April 22nd 2013, 05:00 PMPlatoRe: Functions Sets
- April 23rd 2013, 07:42 AMjohngRe: 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.

Attachment 28106 - April 23rd 2013, 11:08 AMemakarovRe: 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 .