Why Is \left[\frac{1}{2},1\right] a Neighborhood of 1 in \left[0,1\right]?

[funcalys]

Hi folks, as I was reviewing the metric space section in Amann- Escher textbook, I came across the following example of neighborhood:
"For $\left[0,1\right]$ with the metric induced from $R$, $\left[\frac{1}{2},1\right]$ is a neighborhood of 1, but not of $\frac{1}{2}$."
However I can't point out the exactly "r">0 satisfying $B_{[0,1]}(1,r)$$\subseteq[0,1]$.

 

[haruspex]

Won't any r < 1/2 do?

 

[elgen]

[1/2, 1] is a neighborhood of 1. In this case, r=1/2. Any element, of the ball with a radius of 1/2 centered at 1, has a distance less than 1/2 from 1.

[1/2,1] is not a neighborhood of 1/2. This is because any ball with a radius of r>0 centered at 1/2 contains some elements that are not in [1/2, 1].

 

[funcalys]

guess I misunderstood some of the concept in the first place, I thought the ball centered at 1 must completely lie in the interval [1/2,1].
:D.
Thank guys.

 

[HallsofIvy]

It does! What makes you think there are any numbers in [1/2, 1] that are not in [1/2, 1]?

 

[Robert1986]

guess I misunderstood some of the concept in the first place, I thought the ball centered at 1 must completely lie in the interval [1/2,1].
:D.
Thank guys.

So, for this example, we are not concerned with the entire real line, just the closed unit interval. So, I think that you are probably considering points like 1.1 and 1.2 (for example) to be lying in this ball. However, for this example you can just think about those points as not existing because we only care about points in the closed unit interval.