Neighborhood Proof for Distinct Points in a Metric Space

[Tokipin]

Homework Statement

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

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

2. The attempt at a solution

OK, intuitively I recognize at least two cases: 1) If both points are "separate" such that for $a$ and $b$, their smallest neighborhoods are $\{a\}$ and $\{b\}$ 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 $r$ about the point we can find another open ball of radius $r-\epsilon$ about the same point which is a strict subset of the first ball, then we can find a $\delta$ such that the open ball of radius $\delta$ 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.

 

[John Creighto]

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.

 

[Tokipin]

That's hilarious. I was thinking that if we define an open ball of radius $d(a,b)/2$ about each point, then the neighborhoods wouldn't intersect. But I wasn't sure I could generalize this from $R^2$. Thanks.

Here's an attempt at a proof:

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

$$d(a,p) < d(a,b)/2$$
$$d(b,p) < d(a,b)/2$$

This would contradict the triangle inequality of a metric space:

$$d(a,b) \leq d(a,p) + d(p,b)$$

Since each term on the right would be less than $d(a,b) / 2$ and hence their sum would be less than the term on the left.