Basic Proof

**Aryth** · August 29th 2011, 06:52 PM

Let $x$ and $y$ be distinct real numbers. Prove that there is a neighborhood $P$ of $x$ and a neighborhood $Q$ of $y$ such that $P \cap Q = \emptyset$ .

The book left this proof as an exercise and the instructor said nothing about it. I'm just curious as to what the proof might be.

**TKHunny** · August 29th 2011, 07:16 PM

Have you considered a neighborhood smaller than ½|P-Q|?

**Aryth** · August 30th 2011, 08:44 AM

I have tried to consider your suggestion, and it looks similar to the epsilon we chose when proving that a sequence can't converge to two numbers unless they were equal. However, I'm not quite sure what I'm supposed to do with that... We just assumed the OP was true and moved on with it.

P.S. I don't have much experience proving things so it's taking me awhile to think things through.

**Plato** · August 30th 2011, 08:52 AM

Originally Posted by Aryth

I have tried to consider your suggestion, and it looks similar to the epsilon we chose when proving that a sequence can't converge to two numbers unless they were equal. However, I'm not quite sure what I'm supposed to do with that... We just assumed the OP was true and moved on with it.
P.S. I don't have much experience proving things so it's taking me awhile to think things through.

Let $\delta=\frac{|x-y|}{4}>0$ .
Let $P=(x-\delta,x+\delta)~\&~P=(y-\delta,y+\delta)$ .
Show that $P\cap Q=\emptyset$ .

**Aryth** · August 30th 2011, 09:45 AM

Just curious, why did you pick that as $\delta$ ?

**Plato** · August 30th 2011, 09:52 AM

Originally Posted by Aryth

Just curious, why did you pick that as $\delta$ ?

Because it works.

**Aryth** · August 30th 2011, 09:55 AM

That's basically the same thing my instructor said. To me it just seems random, but it does work.

I appreciate your help.

**HallsofIvy** · August 30th 2011, 10:08 AM

The point is that if a specific $\delta$ works, then so does any positive number less than $\delta$ .

TKHunny suggested $\deltas= \frac{|P- Q|}{2}$ which is the largest number that would work so that $N_P(\delta)$ and $N_Q(\delta)$ are disjoint because $\delta+ \delta= \frac{|P-Q|}{2}+ \frac{|P-Q|}{2}= |P- Q|$ . (Draw a picture to see why.) Plato just chose to pick a smaller number. $\delta= \frac{|P-Q|}{3}$ would also work.

And, that is using the same $\delta$ for both sets, which is not necessary. You could also use, say, $\frac{|P-Q|}{5}$ for one open set and $\frac{|P-Q|}{3}$ for the other- just use two numbers that add up to less than |P-Q|.

**Aryth** · August 30th 2011, 10:26 AM

Ah, that makes it clearer. Thank you very much.

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