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Let X be a metric space, p ∈ X, and let K ⊂ X be compact. Show that there exist x0, x1 ∈ K such that d(x0, p) ≤ d(x, p), ∀ x ∈ K, d(x1, p) ≥ d(x, p), ∀ x ∈ K.

I know that I have to show the distance function is contunous and since its domain is compact, the range of a contunous function is compact. Thus, it has a minimum and maximum because it's closed. But not sure how to construct the proof.

user320863
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  • Ostensibly the Question concerns existence of points of minimum and maximum distance in compact subset $K$ from fixed point $p\in X$. If the solution has been reduced to proving continuity of the distance function, then the earlier Question noted by @CarstenS should provide the missing step. – hardmath Mar 08 '16 at 15:12

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This follows simply from the triangle inequality. Let $x,y \in K$ be such that $d(x,y) < \epsilon$. Then, by the reverse triangle inequality,

$$|d(x,p)-d(y,p)| \leq d(x,y) < \epsilon$$

so the metric is continuous and thus attains it's min/max.

Anthony Peter
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