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.