Part CCSIR NET December 2025disjoint-closed-sets-in-a-compact-space-need-not-let-the-summed-distance-reach-0-anywhere

Disjoint closed sets in a compact space need not let the summed distance reach 0 anywhere

Let (X,d) be a metric space. For a non-empty subset A of X, and x∈X, define d(x,A) = $inf_{a\in{}A}$ d(x,a). Which of the following statements are necessarily true?

A.For all x,y∈X and every non-empty subset A of X, we have d(x,A)−d(y,A) ≤ d(x,y).✓
B.For every non-empty subset A of X, the function x↦d(x,A) is uniformly continuous on X.✓
C.A non-empty subset A of X is closed if and only if d(x,A)>0 for all x in X outside A.✓
D.If X is compact and $C_{1},\dots,C_{k}$ are non-empty closed sets of X such that $\cap{}C_{i}=\emptyset$ , then the minimum value of $x\mapsto\sum{}d(x,C_{i})$ on X is 0.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: An arbitrary intersection of open sets is open

Disjoint closed sets in a compact space need not let the summed distance reach 0 anywhere

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