# Prove that elements in a neighborhood have their own neighborhood?

Sorry for the abbreviation above; character limits, and all. Anyway:

1. Homework Statement

"Suppose that ##x∈ℝ## and ##\epsilon>0##. Prove that ##(x-\epsilon,x+\epsilon)## is a neighborhood of each of its members. In other words, if ##y∈(x-\epsilon,x+\epsilon)##, then prove that there is a ##\delta>0## such that ##y∈(y-\delta,y+\delta)⊆(x-\epsilon,x+\epsilon)##."

## Homework Equations

Neighborhood: "A set ##O⊂ℝ## is a neighborhood of ##x∈ℝ## if ##O## contains an interval of positive length centered at ##x##. In other words, there exists an ##\epsilon>0## such that ##(x-\epsilon,x+\epsilon)⊆O##."

## The Attempt at a Solution

Okay, so I get the feeling that I was supposed to do something else for this proof; specifically something I learned from this class. Here is the proof I wrote, otherwise:

"We start with the fact that ##x-\epsilon<y<x+\epsilon##. Subtracting ##x## from the inequality gives ##-\epsilon<y-x<\epsilon##, or ##|y-x|<\epsilon##. Then subtracting ##|y-x|## gives ##0<\epsilon-|y-x|##. Now we set ##0<\delta≤\epsilon-|y-x|##.

Now, we show that the interval ##(y-\delta,y+\delta)## is a non-empty subset of ##(x-\epsilon,x+\epsilon)##. We start with the inequality ##0<\delta≤\epsilon-|y-x|##.
- Adding ##y## gives ##y<y+\delta≤\epsilon-|y-x|+y##.
- Reversing the signs of the inequality and adding ##y## gives: ##y>y-\delta≥y-(\epsilon-|y-x|)##.

We take ##y-x≥0##. The case for if ##x-y≥0## is similar.

- Then the first inequality becomes: ##y<y+\delta≤\epsilon-|y-x|+y=\epsilon-(y-x)+y=\epsilon-y+x+y=x+\epsilon##.
- And then the second inequality becomes ##y>y-\delta≥y-(\epsilon-|y-x|)=y-(\epsilon-(y-x))=y-(\epsilon-y+x)=2y-(x+\epsilon)=(2y-2x)+x-\epsilon=2(y-x)+(x-\epsilon)≥x-\epsilon##

And so combining these statements yields: ##x-\epsilon≤y-\delta<y<y+\delta≤x+\epsilon##. And so there exists a ##\delta>0## such that if ##y## is an element of a neighborhood of ##x##, then there exists a neighborhood of ##y## such that ##(y-\delta,y+\delta)⊆(x-\epsilon,x+\epsilon)##."

I have a strong feeling that I was supposed to do something else, but cannot think of what it could be.

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## Answers and Replies

mfb
Mentor
(I extended the title)

That works. You can do it more compact with the triangle inequality but analyzing the separate cases works as well.
Now we set ##0<\delta≤\epsilon-|y-x|##
I would write that as "now we choose a ##\delta## such that..." to make clear what you set. Alternatively you can explicitly define one (e.g. the average between the two sides).

Gotcha. Thanks.