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Basic Neighborhood Proof

  • Thread starter Tokipin
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1. Homework Statement

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.
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.
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.

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