1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Neighborhood Proof

  1. Mar 11, 2009 #1
    1. The problem statement, all variables and given/known data

    From Introduction to Topology by Bert Mendelson, Chapter 2.4, Exercise 6:

    Let [itex]a[/itex] and [itex]b[/itex] be distinct points of a metric space X. Prove that there are neighborhoods [itex]N_a[/itex] and [itex]N_b[/itex] of [itex]a[/itex] and [itex]b[/itex] respectively such that [itex]N_a \cap N_b = \varnothing[/itex].

    2. The attempt at a solution

    OK, intuitively I recognize at least two cases: 1) If both points are "separate" such that for [itex]a[/itex] and [itex]b[/itex], their smallest neighborhoods are [itex]\{a\}[/itex] and [itex]\{b\}[/itex] respectively, then these neighborhoods obviously don't intersect.

    2) If at least one of the points has an "infinitely partitionable neighborhood" such that for any open ball of radius [itex]r[/itex] about the point we can find another open ball of radius [itex]r-\epsilon[/itex] about the same point which is a strict subset of the first ball, then we can find a [itex]\delta[/itex] such that the open ball of radius [itex]\delta[/itex] about that point does not include a neighborhood of the other point.

    Where I think I'm getting confused is whether these are the only two scenarios that can occur in a metric space (both points separate, or one or both possessing "infinitely partitionable neighborhoods.")

    I was trying to find a contradiction that might arise due to the symmetry of the distance function, but I can't find one. I might also be thinking too much into it, so overall I'm quite confused.
     
  2. jcsd
  3. Mar 11, 2009 #2
    I think all you need to do is define your neighborhood so all points within it are less then half the distance between the two points.
     
  4. Mar 12, 2009 #3
    That's hilarious. I was thinking that if we define an open ball of radius [itex]d(a,b)/2[/itex] about each point, then the neighborhoods wouldn't intersect. But I wasn't sure I could generalize this from [itex]R^2[/itex]. Thanks.

    Here's an attempt at a proof:

    About each point [itex]a[/itex] and [itex]b[/itex], define an open ball of radius [itex]d(a,b)/2[/itex]. If there was a point [itex]p[/itex] which was in both of the open balls, then this would mean that both of the following would be true:

    [tex]d(a,p) < d(a,b)/2[/tex]
    [tex]d(b,p) < d(a,b)/2[/tex]

    This would contradict the triangle inequality of a metric space:

    [tex]d(a,b) \leq d(a,p) + d(p,b)[/tex]

    Since each term on the right would be less than [itex]d(a,b) / 2[/itex] and hence their sum would be less than the term on the left.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Basic Neighborhood Proof
  1. Basic Proof (Replies: 4)

Loading...