Recognitions:
Gold Member

 SETS DEFINED BY SPECIFICATION: Given a set X and a sentence P(x) that is either true or false whenever x is any particular element of X, there is a set whose elements are precisely those x ∈ X for which P(x) is true, denoted by {x ∈ X : P(x)}.
Does this mean that whenever the function P(x) is true, then x is an element of X, and when P(x) is false, then x is not an element of X?

I'm confused because the wording says that "...a sentence P(x) that is either true or false whenever x is any particular element of X..." which leads me to believe that whether P(x) is either true or false, then it is still an element of set X.

Or is it saying that there is a set within X in which P(x) is true, and there is also another set within X in which P(x) is false?
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Mentor Blog Entries: 8 Hmm, no, it's neither of these. The set $\{x\in X~\vert~P(x)\}$ are all the elements of X which satisfy P. So, in particular, we want all elements to lie in X. Maybe some examples may help you to see the concept: $A=\{x\in \mathbb{N}~\vert~x~\text{is even}\}$ This means that we take all the elements of $\mathbb{Z}$ which are even. So the elements of our set A are precisely 0,2,4,6,8,10,12,... $B=\{x\in \mathbb{Z}~\vert~x>0\}$ This is the set of all positive integers. The set consists out of 1,2,3,4,... Note that x=1/2 also satisfies x>0. But 1/2 does not belong to B because it doesn't satisfy $x\in \mathbb{Z}$. $C=\{x\in \mathbb{Q}~\vert~2x+2=0\}$ This is the set of all rational numbers x such that 2x+2=0. The only number satisfying that is of course -1. And this number is in $\mathbb{Q}$, so it belongs to C. Does this clear up some things?
 Recognitions: Gold Member In those 3 examples, would A be considered a proper subset of N, B a proper subset of Z, and C a proper subset of Q?

Mentor
Blog Entries: 8

$$D=\{x\in \mathbb{N}~\vert~x\geq 0\}$$
So this set consists of all elements in $\mathbb{N}$ which are larger than 0. Clearly, this is the entire set. So D is not a proper subset in this case.