Neighborhoods in Advanced Calculus I

[Grant Bonner]

Assume x, y are distinct real numbers. Prove that the neighborhood P of x and the neighborhood Q of y do not intersect. i.e. (P intersect Q)= Empty set.

The only thing I have come up with is to assume there is some positive distance between x and y, I call it n, where n= |x-y|. Making that assumtion, and by the definition of neighborhoods of P (x - e, x + e: centered at x) and Q (y - d, y + d: centered at y), I say that in order for (P intersect Q)= Empty set, e,d < (n/2).

It obviously looks prettier on paper, but am I heading in the correct direction here? I think what is throwing me off is that when told x,y are DISTINCT real numbers.

I'm new to the math forum, and may not use the correct syntax. If so, sorry! Thanks.

Grant

 

[pka]

Assume x, y are distinct real numbers. Prove that the neighborhood P of x and the neighborhood Q of y do not intersect. i.e. (P intersect Q)= Empty set.

First of all that is not true as stated.
It should: Prove that there is a neighborhood P of x and there is neighborhood Q of y that do not intersect.
Because x, y are distinct real numbers, let \(\displaystyle \varepsilon = \frac{{\left| {x - y} \right|}}{4} > 0\).
Now let \(\displaystyle P = \left( {x - \varepsilon ,x + \varepsilon } \right)\;\& \,Q = \left( {y - \varepsilon ,y + \varepsilon } \right)\).

Now you finish it off.

 

[Grant Bonner]

Why shouldn't we let e < |x-y|/2? |x-y|/4 (or any positive, real number) will always be > 0. Regardless, I'm not seeing where to go from here. I'm fairly new to proofs and as they are not truly "defined," I struggle with them a bit.

 

[pka]

Why shouldn't we let e < |x-y|/2? |x-y|/4 (or any positive, real number) will always be > 0. Regardless, I'm not seeing where to go from here. I'm fairly new to proofs and as they are not truly "defined," I struggle with them a bit.

Suppose that \(\displaystyle z\in P\cap Q.\)
Then by definition \(\displaystyle \left| {x - z} \right| < \varepsilon \;\& \,\left| {z - y} \right| < \varepsilon\).
Then we get a contradiction.
\(\displaystyle \left| {x - y} \right| \leqslant \left| {x - z} \right| + \left| {z - y} \right| < 2\varepsilon = \frac{{\left| {x - y} \right|}}{2}\)

 

[Grant Bonner]

Gotcha. Again though, shouldn't we let 0 < e < |x-y|/2 ? So that 2e < |x-y|, as to have the equation |x-y|<=|x-z|+|z-y|<|x-y| to get our contradiction? Haha, sorry if I'm bothersome, I'm only looking to understand the solution. I've spent 4 years in college and this is the first subject that I cannot fully grasp.

 

[pka]

Gotcha. Again though, shouldn't we let 0 < e < |x-y|/2 ? So that 2e < |x-y|, as to have the equation |x-y|<=|x-z|+|z-y|<|x-y| to get our contradiction? Haha, sorry if I'm bothersome, I'm only looking to understand the solution. I've spent 4 years in college and this is the first subject that I cannot fully grasp.

If you are asking if we could have used \(\displaystyle \varepsilon = \frac{{\left| {x - y} \right|}}{2}?\) The answer is yes.

We could also have used \(\displaystyle \varepsilon = \frac{{\left| {x - y} \right|}}{200}\).

 

[Grant Bonner]

well, e < |x-y|/2. Assuming x was 1 and y was 2, the neighborhoods would intersect at 3/2 if e=|x-y|/2. But thanks alot for the help!

 

[pka]

well, e < |x-y|/2. Assuming x was 1 and y was 2, the neighborhoods would intersect at 3/2 if e=|x-y|/2. But thanks alot for the help!

No they would not
\(\displaystyle P = \left( {\frac{1}{2},\frac{3}{2}} \right)\;,\,Q = \left( {\frac{3}{2},\frac{5}{2}} \right)\;\& \,P \cap Q = \emptyset\)

 

[Grant Bonner]

Word.

 

[mmm4444bot]

Word.

Sentence is better. :wink: