Claim: For all integers n>1, a set C ⊆ R^n is compact if and only if C is closed and bounded.

For the forward direction, suppose C is compact. Choose some a in R^n and let B(a, k) be an open ball centered at a of radius k. Note that B(a, 1) ∪ B(a, 2) ∪ ... = R^n since for all b in R^n, d(a, b) < k for some integer k as a result of the Archimedean property. Therefore, { B(a, k) }_k is an open cover of C. By compactness, there exists a finite subcover { B(a, k_1), B(a, k_2), ..., B(a, k_N) }. Define K := max {k_1, k_2, ..., k_N}. Thus, C ⊆ B(a, K), and by definition, C is bounded.

Note that because R^n is a T_1 space, if C is finite, then C is closed. Suppose now that C is infinite. Let a be some fixed point not in C and let b be some point in C. Since R^n is T_2, there exist disjoint open sets U_b and V_b such that a ∈ U_b and b ∈ V_b. Since { V_b }_b is an open cover of C, there exists a finite subcover { V_b_1, V_b_2, ..., V_b_M }. Define U_a := U_b_1 ∩ U_b_2 ∩ ... ∩ U_b_M. Clearly, for all b ∈ C, b ∉ U_a, meaning that U_a ⊆ R^n ∖ C for all a. Therefore, ∪_a U_a = R^n ∖ C. This shows R^n ∖ C is an open set and hence that C is closed.

For the backward direction, suppose now that C is closed and bounded. Since C is bounded, there exists some H>0 such that C ⊆ [-H, H]^n. First, we shall prove the compactness of [-H, H]. Consider an open cover U = { U_i } of [-H, H] and define V := { x ∈ [-H, H] | [x, H] is covered by finitely many elements in U }. Define v := inf V, which exists since V is bounded below by -H. There exists some U_H ∈ U such that H ∈ U_H. Thus, for some ε>0, (H-ε, H] ⊆ (H-ε, H+ε) ⊆ U_H. Therefore, for all x ∈ (H-ε, H], x ∈ V, proving that v < H. Similarly define U_v ∈ U such that v ∈ U_v. There exists some 𝛿>0 such that [v, v+𝛿) ⊆ U_v. Since v = inf V, there exists some w ∈ V such that w ∈ [v, v+𝛿). This implies that [v, H] = [v, v+𝛿) ∪ [w, H], so v ∈ V and hence v=-H. This proves that [-H, H] is compact. By identifying R^n with the product topology on R, [-H, H]^n is trivially compact.

Finally, consider any open cover A of C and define B := A ∪ { R^n ∖ C } . Since C is closed, R^n ∖ C is open, making B an open cover of [-H, H]^n. By compactness, there exists a finite open cover B' of [-H, H]. Now define A' := B' ∖ { R^n ∖ C }. For all c ∈ C, c is in an open set in B' that is not R^n ∖ C. In short, c is in an open set in A', making A' a finite open cover for A. Therefore, C is compact. ■

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