Undergrad Open balls dense in closed balls in Euclidean space

Click For Summary
The discussion centers on the confusion regarding the statement that open balls are dense in closed balls in Euclidean space, specifically in ##\mathbb{R}^n##. It clarifies that while this statement is true in ##\mathbb{R}^n## with the standard Euclidean metric, it does not apply to all metric spaces. The mention of discrete metrics serves as a counterexample to the general claim about open and closed balls. The participants emphasize that the standard metric in ##\mathbb{R}^n## is derived from an inner product, which leads to a specific understanding of density. Overall, the conversation highlights the nuances of metric spaces and their implications for the density of open and closed balls.
psie
Messages
315
Reaction score
40
TL;DR
I am working a problem in Gamelin and Greene's book on topology. They ask about whether closed balls are closed sets (which they are), but moreover if the closure of an open ball is a closed ball. They make a statement concerning this which I don't understand.
Any set with at least two elements and equipped with the discrete metric is a counterexample to the claim that the closure of an open ball is a closed ball. Yet, in the back of the back book where they present solutions to some of their exercises, they write:

The statement about open balls being dense in closed balls holds in ##\mathbb R^n##, but it does not hold in metric spaces in general.

I feel silly for asking, but I can not make sense logically of the first part of the sentence. First they say it holds in ##\mathbb R^n##, but not in metric spaces in general. What could they mean by it holds in ##\mathbb R^n##? My understanding is that the statement holds in ##\mathbb R^n## equipped with a metric derived from a norm, but not otherwise. Is this correct?
 
Physics news on Phys.org
\mathbb{R}^n, without more, means \mathbb{R}^n with the euclidean norm.
 
  • Like
Likes psie and FactChecker
Somewhat related: The standard metric in ##\mathbb R^n## is derived from an inner-product## <,>##, which gives rise to a norm ##||.||##, through , ##||v||=<v,v>^{1/2}##, and a metric ##m(x,y):=<x-y, x-y>^{1/2}##
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K