# Thread: Writing a Proof using Def of neighborhood

1. ## Writing a Proof using Def of neighborhood

Ok so I'm really struggling in my analysis class and could really use some help along with some in depth explanation.

So my problem says:
Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x + epsilon) is a neighborhood of each of its members; In other words if y exists in (x - epsilon, x + epsilon) then there is rho > 0 such that (y - rho, y + rho) is a subset of (x - epsilon, x + epsilon)

I know the second part pretty much gives it away. I just have the hardest time writing proofs. Your help is GREATLY appreciated!!!

2. ## Re: Writing a Proof using Def of neighborhood

Originally Posted by arlayton
In other words if y exists in (x - epsilon, x + epsilon) then there is rho > 0 such that (y - rho, y + rho) is a subset of (x - epsilon, x + epsilon)
As stated by you here all you need to do is determine a suitable $\displaystyle \rho$. Can you think of any way of determining a suitable real number $\displaystyle \rho < \epsilon$.
Spoiler:
Now given $\displaystyle \epsilon > 0, \exists n \in \mathbb{N} \ni n.\epsilon > 1 \implies \epsilon > \frac{1}{n}$ by Archimedian Property. Now choose $\displaystyle \rho = \frac{1}{n}$ by definition $\displaystyle (x-\rho,x+\rho) = \{ y \in \mathbb{R} | x-\rho < y < x+\rho \} \subset (x-\epsilon,x+\epsilon)$

3. ## Re: Writing a Proof using Def of neighborhood

Originally Posted by kalyanram
As stated by you here all you need to do is determine a suitable $\displaystyle \rho$. Can you think of any way of determining a suitable real number $\displaystyle \rho < \epsilon$.
Spoiler:
Now given $\displaystyle \epsilon > 0, \exists n \in \mathbb{N} \ni n.\epsilon > 1 \implies \epsilon > \frac{1}{n}$ by Archimedian Property. Now choose $\displaystyle \rho = \frac{1}{n}$ by definition $\displaystyle (x-\rho,x+\rho) = \{ y \in \mathbb{R} | x-\rho < y < x+\rho \} \subset (x-\epsilon,x+\epsilon)$
Um, there's nothing wrong with what you did, but I don't think it answers the question. You basically showed that (x-a,x+a) is contained in (x-b,x+b) if b>a.

Originally Posted by arlayton
In other words if y exists in (x - epsilon, x + epsilon) then there is rho > 0 such that (y - rho, y + rho) is a subset of (x - epsilon, x + epsilon)
As you said, the second part gives it away, but you're having difficulty writing the proof.

Start with the "if" stuff. In this case, you want a "y" that's in (x-e,x+e).
Then manipulate it (using "theorems", axioms, or definitions) to get the conclusion. In this case, you're looking for an "r" so that (y-r,y+r) is a subset of (x-e,x+e).
Can you think of a way to get this "r"?
How far is y from x-e? (let's say it's "a", i.e. y-(x-e)=a)
How far is y from x+e? (let's say it's "b", i.e. x+e-y=b)
Is (y-a,y+a) a subset of (x-e,x+e)? How about for (y-(a/2),y+(a/2))?
Is (y-b,y+b) a subset of (x-e,x+e)? How about for (y-(b/2),y+(b/2))?
Does "r" depend on how close "y" is to x-e or x+e? (If you've encountered the min function, that would make things easier)

4. ## Re: Writing a Proof using Def of neighborhood

Originally Posted by arlayton
So my problem says:
Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x + epsilon) is a neighborhood of each of its members; In other words if y exists in (x - epsilon, x + epsilon) then there is rho > 0 such that (y - rho, y + rho) is a subset of (x - epsilon, x + epsilon)
Suppose that $\displaystyle \varepsilon > 0~\&~t \in \left( {x - \varepsilon ,x + \varepsilon } \right)$

Let $\displaystyle \delta = \min \left\{ {\frac{{t - x + \varepsilon }}{2},\frac{{x + \varepsilon - t}}{2}} \right\}$ the show that $\displaystyle \left( {t - \delta ,t + \delta } \right) \subseteq \left( {x - \varepsilon ,x + \varepsilon } \right)$.

5. ## Re: Writing a Proof using Def of neighborhood

1. You often really have two different activities. You'll often work backward, assuming the result, and then try to find the rho or delta or whatever that works. But that work isn't part of your proof. Your proof should work forwards, where you just present the correct rho, and then show that it actually works to solve the problem. And although the "forward proof" will often look like it's just copying your work backwards, that isn't *always* the case. (Ex: if working backward you invoke ".. and since x<3, we know that x<5...", you won't be able to just mindlessly copy that reasoning when going "the other direction" - saying "... and since x<5, we know that x<3..." is wrong.) You'll still have to think about how to structure the forward demonstration, even if you know you've gotten the rho that "works". So it's like having to do two problems.
2. Be as systematic as possible. Take little steps, following the definitions or theorems as needed to justify your claims.
3. Be structured. Make clear any assumptions or notations that might not be obvious. Especially for long proofs, break down your steps into lemmas, claims, etc. Start new paragraphs when starting different parts of the task. Indicate what you're going to do, do it, and then summarize what you did. ("Proof by contraction: ASSUME there is no such x. Then... ... But that contradicts how y was choosen. Thus the intial assumption was false. Therefore such an x exists, meaning ...") ("Induction: P(1) holds, because... . Now assume P(k) holds for some k>=1. Then... ... and so P(k+1) holds. Therefore, by induction, P(n) holds for all n>=1.)
4. In this case:
Spoiler:

For any real x, and real e>0, define U = (x-e,x+e).
Let y in U. Will prove that U is a neighborhood of y.
Given y, let rho = min( |y-(x-e)|, |y-(x+e)| ). Note that rho > 0, because... . Let V = (y-rho, y+rho). Clearly V is an open neighborhood y.
Claim V is a subset of U:
Let z in V. Then.... and so z>x-e. Also, from ... get that z<x+e. Therefore z in U, proving that V is a subset of U.
Thus have shown that any element y of U is contained in an open neighborhood V that is entirely contained in U, and hence that U is neighborhood of each of its elements.

6. ## Re: Writing a Proof using Def of neighborhood

Plato's post *is* the answer, but how did he arrive at it?

what we have is something like this:

x-ε......x......x+ε

so when we have t in (x-ε,x+ε), we have either:

x-ε...t..x.....x+ε

or:

x-ε.....x..t...x+ε

in the first case, where t is closer to x-ε than x+ε, if we take "half the distance between t and x-ε", we're SURE to be inside (x-ε,x+ε).

that "half distance" is (1/2)(t - (x-ε)) = (1/2)(t - x + ε), which is the first of Plato's numbers.

the other case, where t is closer to x+ε than x-ε, the distance we want is (1/2)(x+ε - t), which is the second number in Plato's post.

in point of fact, the factor 1/2 isn't really needed, we could take:

$\displaystyle \rho = \min\{t - x + \epsilon,x + \epsilon - t\}$ just as easily or any factor 0 < a ≤ 1 times these two numbers.

note that t = x is a special case, which reduces to $\displaystyle \rho = \epsilon$.

the point is, t can only be ε away from x, which means t can only be at most ε - |x-t| away from one of x+ε or x-ε (|x-t| is less than ε, because t is in the interval
(x-ε,x+ε), see?).

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### proofs epsilon neighbourhood of real numbers

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