Is x just a real number in this case? I am given:
Let a ∈ ℝ and let A ∈ (0,∞) with x ∈ (a - A, a + A). Then there exists some P ∈ (0,A) such that x ∈ (a - P, a + P).
I'm trying to figure this one out.  I see that a is a real number, and A is greater than or equal to zero.  Does that mean with this notation that x is just a real number, since it is an element of a minus all elements greater than or equal to zero and also greater than?
 A: Yes, $x$ is a real number. Remember that $(a-A,a+A)$ is the notation for an open interval; hence $(a-A,a+A)\subset\mathbb{R}$. Therefore, since $x\in(a-A,a+A)$, $x\in\mathbb{R}$ by the definition of a subset.
A: Yes, $x$ is a real number. But it's not any real number. By the time $x$ is identified, the number $A$ has already got a value. Suppose, for instance, that $a = 7$, and $A = 2$. Then $x$ is not just any real number ... it's a real number between $7-2$ and $7+2$, i.e., 
$$
5 < x < 9.
$$
The claim is that if you have such a number -- say $x = 5.1$ -- then there's a number $P$, less than $A$, with the property that $x$ lies between $a - P$ and $a + P$. In this example, the number $P = 1.95$ would work, because $5.1$ lies between $7 - 1.95 = 5.05$ and $7 + 1.95 = 8.95$. 
In short, if $x$ lies in an open interval around $a$ of width $4$, then it also lies in an open interval of some width a little less than $4$. 
Note that if we were talking about closed intervals, we'd have
$$
5 \le x \le 9
$$
and $x = 5$ is a possible value. But then you cannot reduce the size of the interval. So the open-ness is essential to what you're trying to show. 
A: This is saying that if $x$ lies in the open ball of radius $A$ centered at $a$, then it also lies in an open ball with the same center but smaller radius. The crucial thing here is that they are open balls--it wouldn't be true if these were closed balls.
