How can we show K is compact?

Let X be a compact subset of R^n and r > 0. Show that K = U (x ∈ X) C(x, r) is compact, where C(x, r) is the closed ball of radius r centered at x.

Thank you.

3 Answers

  • 7 years ago
    Favorite Answer

    Let (k_n) be a sequence in K. Since each k_n is in at least one of the balls C(x, r), there is a sequence (x_n) in X such that ||k_n - x_n|| ≤ r for all n (1). Since X is compact, (x_n) has a subsequence (x_n_m) that converges to some x ∈ X. Then, (x_n_m) is bounded and, in virtue of (1), (k_n_m) is bounded, too. So, it follows from Bolzano Weierstrass theorem for sequences that (k_n_m) has a subsequence that converges to some k ∈ R^n.

    Renumbering the terms of this subsequence of (k_n_m) and picking the corresponding subsequence of (x_n_m), we see there are subsequences (k_i) of (k_n) and (x_i) of (x_n) such that

    k _i → k ∈ R^n

    x_i → x ∈ X

    ||k_i - x_i|| ≤ r for all i

    From these 3 conditions it follows that ||k - x|| ≤ r, which implies that k ∈ C(x, r) ⊂ K. Hence, k ∈ K and we conclude every sequence in K has a subsequence that converges in K. As we know, this implies K is compact.

  • jibz
    Lv 6
    7 years ago

    Since X is bounded, X ⊆ C(z,ρ) for some z ∈ ℝ^n and ρ ∈ ℝ. To show that K is bounded, let k ∈ K be given. Then k ∈ C(x,r) for some x ∈ X. Hence


    ≤ |k-x| + |x-z| via the triangle inequality,

    ≤ r + ρ.

    Since the above holds for all k ∈ K, we have K ⊆ C(z, r + ρ), i.e. K is bounded.

    Next, we'll show that K is closed by showing that its complement ℝ^n \ K is open. So let q ∈ ℝ^n \ K be given. Let f : ℝ^n → ℝ be given by f(x) = |x-q|. Since f is continuous, it maps compact sets to compact sets. In particular, f[X] is a compact subset of ℝ. So f[X] has a minimal element m ≥ 0. Note that since q ∈ ℝ^n \ K, we have

    C(q,r) ⊆ ℝ^n \ X,

    which means that actually m > r. We claim that the open ball B(q,m-r) centered at q with radius m-r is a subset of ℝ^n \ K. Ad absurdum, suppose otherwise. Then there exists k ∈ K such that |q-k| < m - r, and, since k ∈ K, there exists x ∈ X be such that |k-x| ≤ r. But then


    ≤ |q-k| + |k-x| via the triangle inequality,

    < m - r + r

    = m,

    contradicting that f[X] ⊆ [m; ∞). This proves the claim. Consequently, q is an interior point of ℝ^n \ K. Since q was arbitrary, we've shown that ℝ^n \ K is open, so K is closed.

    Thus, since K is closed and bounded, K is compact.

  • 7 years ago

    I'm assuming you know that a closed, bounded set in R^n is compact. (Borel-Weierstrauss Thm, I believe?)

    Each of the C(x, r) is closed & bounded.

    The union of closed sets is a closed set.

    K = union of closed, bounded sets, so it is also closed & bounded. Therefore, it is compact.

Still have questions? Get your answers by asking now.