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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?

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You should regard $A$ and $P$ as positive distances, and you should regard $a$ and $x$ as points on the real line. The first sentence says "Suppose the point $x$ is less than $A$ units away from the point $a$". The second sentence says "Then the point $x$ is actually less than $P$ units away from $a$ for some number $P$ smaller than $A$". –  MPW Jul 7 at 0:00

3 Answers 3

up vote 5 down vote accepted

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.

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Thanks! I'm guessing then since P ∈ (0,A) and x ∈ (a - A, a + A), that there is NO x ∈ (a - P, a + P)? –  westhe32nd Jul 6 at 20:36
    
@westhe32nd If $x\in (a-P,a+P)$ then $x\in(a-A,a+A)$, but the converse generally does not hold true. Depends on your choice of $x$. Eg., let $a=0$, $A=2$ and $P=1$. –  Sanath Jul 6 at 20:37
    
This doesn't answer the question. What the OP is trying to show is that any open interval is... open. –  Pedro Tamaroff Jul 6 at 20:54
    
@PedroTamaroff I understand what the OP is trying to show. But the OP's question is as follows: "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?" –  Sanath Jul 6 at 20:55

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.

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So then, by definition, a>A? Since the interval is up, that makes sense. It would seem then that there IS x ∈ (a - P, a + P) since we only have to show one case where this is true? –  westhe32nd Jul 6 at 20:42
    
@westhe32nd No, $a$ need not be greater then $A$. –  Sanath Jul 6 at 20:43
    
@westhe32nd If you are after a proof you need to show that a P exists for all real a, positive A and x in (a-A,a+A), to do so you can consider P=|x|+(A-|x|)/2. –  Taemyr Jul 7 at 6:37

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.

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