- #1

Eclair_de_XII

- 1,062

- 89

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

"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)##."

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.

1. Homework Statement1. 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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