# Partially ordered set

1. Nov 19, 2007

### ar6

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I am quite confused as to how to define the function. Any help would be appreciated.

2. Nov 19, 2007

### varygoode

The new set, $$f(a)$$, is a set consisting of all $$x \epsilon A$$ s.t. $$x \leq a$$.

For example, consider $$A = \mathbb{N}$$. Look at $$f(5)$$. This is simply $$\{1,2,3,4,5\}$$.

So, for the first part, you need to show that

$$f(a)=f(b) \Rightarrow a=b$$

I'll stop here, just in case you need more assistance.

Last edited: Nov 19, 2007
3. Nov 20, 2007

### HallsofIvy

Staff Emeritus
In other words, f(a) is the set of all members of A that are less than or equal to a.

If A were the set of real numbers with the usual order (which is, or course, not "partial"), then f(3)= ($-\infty$,3]. If A were a collection of sets, with "<" meaning set inclusion (that is a "partial" order) then f(a) is the collection of all subsets of a.

4. Nov 20, 2007

### Hurkyl

Staff Emeritus
What's wrong with $f(a) = \{ \, x \mid x \in A \wedge x \leq a \, \}$?

5. Nov 20, 2007

### ar6

Yea I was confused last night and reading too much into the question. I got it (i think) now.