Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 2 / 2G / viii / Solution

True. If there were only finitely many isolated points, delete all of them. The remainder would be a nonempty closed, countable, complete subspace, so part (v) would give it an isolated point. Its positive distance from the finite deleted set would make it isolated in the original space as well, a contradiction. Thus the stronger isolated points in a countable complete metric space result applies.

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