- #1
Tokipin
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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.