Prove that elements in a neighborhood have their own neighborhood?

[Eclair_de_XII]

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.

 

[mfb]

(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).

 

[Eclair_de_XII]

Gotcha. Thanks.