# Stumped on 2 easy function problems!

1. May 3, 2008

### binks01

The first question I have is simple, but when I attempted it, I got stuck.

I'm trying to prove that if f:X->Y and A & B are subsets of X, that f(A intersect B) is a subset of f(A) intersect f(B).

I started by trying to show set containment, beginning with an arbitrary element in f(A intersect B). However, I cannot figure out how to transition into the right hand side of the problem.

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The second question I have is proving that if A and B are finite sets having the same cardinality and f:A->B is one-to-one then f is onto.

I missed class this day and can't figure out what cardinality is by reading the chapter.

2. May 3, 2008

### mutton

1. If $$y \in f(A \cap B)$$, then $$\exists x \in A \cap B$$ such that f(x) = y. Then...

2. The cardinality of a set is its "size". The cardinality of a finite set is the number of elements in it. Since A and B are finite sets having the same cardinality, |A| = |B| = n for some natural number n.

3. May 3, 2008

### binks01

How do you use |A| = |B| = n to prove f is onto?

4. May 3, 2008

### d_leet

What have you tried so far? What happens if it isn't onto? In other words what happens if there is an element of B that is not mapped to by any element of A?

5. May 4, 2008

### Nedeljko

Precise answer depends on used finite set definition.