Compactness in R^n

We use the nested-interval bisection method, first for $n=1$ and then by iterating across coordinates for general $n$. Case $n = 1$. Suppose $(x_k)$ is a bounded sequence of real numbers, so $x_k \in [a_0, b_0] =: I_0$ for all $k$, for some $a_0 < b_0$. Bisect $I_0$ into two closed halves of equal length. At least one half contains $x_k$ for infinitely many indices $k$ (since the set of all indices is infinite and it splits into the union of the index sets of the two halves — if both were finite their union would be finite, a contradiction). Call that half $I_1 = [a_1,b_1]$, and note $b_1 - a_1 = (b_0-a_0)/2$. Repeat: bisect $I_1$, and choose a half $I_2$ containing $x_k$ for infinitely many $k$, again possible by the same finite-union argument. Inductively construct closed intervals

each containing $x_k$ for infinitely many indices $k$. Since the $I_m$ are nested closed intervals with lengths tending to $0$, the Nested Interval Property of $\mathbb{R}$ (a consequence of completeness) gives a unique point $x^* \in \bigcap_{m=0}^\infty I_m$. Now build the subsequence: choose $k_1$ with $x_{k_1} \in I_1$. Having chosen $k_1 < k_2 < \cdots < k_{m-1}$ with $x_{k_j} \in I_j$, use the fact that $I_m$ contains $x_k$ for infinitely many $k$ to choose $k_m > k_{m-1}$ with $x_{k_m} \in I_m$. This is possible at every stage because "infinitely many $k$" always leaves room to pick an index larger than the finitely many already used. Then for every $m$, both $x_{k_m}$ and $x^*$ lie in $I_m$, so

Hence $x_{k_m} \to x^*$, proving the case $n=1$. Case general $n$. Write $x_k = (x_k^{(1)}, \dots, x_k^{(n)})$. Since $(x_k)$ is bounded in $\mathbb{R}^n$, each coordinate sequence $(x_k^{(i)})_{k=1}^\infty$ is a bounded sequence of real numbers (bounded by the same $M$, since $|x_k^{(i)}| \le \|x_k\| \le M$). Apply the $n=1$ case to the first coordinate: extract a subsequence $(x_{k}^{(1)})_{k \in S_1}$ (indices restricted to an infinite subset $S_1 \subset \mathbb{N}$) converging to some $x^{*(1)} \in \mathbb{R}$. Now restrict attention to indices in $S_1$ and apply the $n=1$ case to the second coordinate sequence $(x_k^{(2)})_{k \in S_1}$, which is still bounded: extract a further infinite subset $S_2 \subset S_1$ along which $x_k^{(2)} \to x^{*(2)}$. Crucially, along $S_2$ the first coordinate still converges to $x^{*(1)}$, because $S_2 \subset S_1$ and every subsequence of a convergent sequence converges to the same limit. Iterate this for coordinates $3, 4, \dots, n$, producing nested infinite index sets

such that along $S_n$, every coordinate $i = 1, \dots, n$ converges: $x_k^{(i)} \to x^{*(i)}$ for $k \in S_n$, $k \to \infty$. List the elements of $S_n$ in increasing order as $k_1 < k_2 < \cdots$; this defines the desired subsequence $(x_{k_j})$. By definition of convergence in $\mathbb{R}^n$ (coordinatewise convergence, equivalent to convergence in the Euclidean norm since

for each $i$), we conclude $x_{k_j} \to x^* = (x^{*(1)}, \dots, x^{*(n)})$ as $j \to \infty$. $\blacksquare$

We prove (a) $\Rightarrow$ (b) $\Rightarrow$ (c) $\Rightarrow$ (a), closing the logical circle. The technique for (a)$\Rightarrow$(b) is direct use of Bolzano–Weierstrass plus closedness; for (b)$\Rightarrow$(c) we use a two-step argument (total boundedness plus a Lebesgue-number-style covering argument); for (c)$\Rightarrow$(a) we exhibit explicit covers that fail to have finite subcovers when closedness or boundedness is dropped.

(a) $\Rightarrow$ (b). Assume $K$ is closed and bounded. Let $(x_k)$ be any sequence with $x_k \in K$ for all $k$. Since $K$ is bounded, $(x_k)$ is a bounded sequence in $\mathbb{R}^n$, so by Bolzano–Weierstrass theorem in $\mathbb{R}^n$ there is a subsequence $x_{k_j} \to x^*$ for some $x^* \in \mathbb{R}^n$. We must show $x^* \in K$. Since $K$ is closed, $\mathbb{R}^n \setminus K$ is open; if $x^*$ were in $\mathbb{R}^n \setminus K$, there would be an open ball $B(x^*, \varepsilon) \subset \mathbb{R}^n \setminus K$. But $x_{k_j} \to x^*$ means eventually $x_{k_j} \in B(x^*,\varepsilon)$, and $x_{k_j} \in K$ for all $j$ — a contradiction, since $B(x^*,\varepsilon)$ is disjoint from $K$. Hence $x^* \in K$, proving sequential compactness.

(b) $\Rightarrow$ (c). Assume $K$ is sequentially compact. We argue in two steps. Step 1 (total boundedness). We claim: for every $\varepsilon > 0$, $K$ can be covered by finitely many open balls of radius $\varepsilon$ centered at points of $K$. Suppose not, for some $\varepsilon_0 > 0$. Then we may construct a sequence $(x_k)$ in $K$ inductively as follows: pick $x_1 \in K$ arbitrarily. Having chosen $x_1, \dots, x_k$, since (by assumption) the balls $B(x_1,\varepsilon_0), \dots, B(x_k,\varepsilon_0)$ do not cover $K$, pick $x_{k+1} \in K \setminus \bigcup_{i=1}^k B(x_i, \varepsilon_0)$. By construction $\|x_i - x_j\| \ge \varepsilon_0$ for all $i \ne j$. Such a sequence has no convergent subsequence, since any subsequence would have all pairwise distances $\ge \varepsilon_0$, so no subsequence is Cauchy, hence none converges. This contradicts sequential compactness of $K$. Hence for every $\varepsilon > 0$, $K$ has a finite $\varepsilon$-net. Step 2 (finite subcover from any open cover). Let $\mathcal{U} = \{U_\alpha\}_{\alpha \in A}$ be an open cover of $K$. Suppose toward contradiction that no finite subcollection of $\mathcal{U}$ covers $K$. By Step 1, for each $m \in \mathbb{N}$, cover $K$ by finitely many balls of radius $1/m$ centered at points of $K$. If every one of these finitely many balls could be covered by finitely many $U_\alpha$'s, then (taking the union of these finite subcollections over all the balls, still a finite union since there are finitely many balls) $K$ itself would be covered by finitely many $U_\alpha$'s, contradicting our assumption. Hence for each $m$, at least one ball $B(x_m, 1/m)$ (with $x_m \in K$) has the property that $B(x_m,1/m) \cap K$ is not covered by any finite subcollection of $\mathcal{U}$. This gives a sequence $(x_m)$ in $K$. By sequential compactness, some subsequence $x_{m_j} \to x^* \in K$. Since $\mathcal{U}$ covers $K$, $x^* \in U_{\alpha_0}$ for some $\alpha_0 \in A$; since $U_{\alpha_0}$ is open, there is $\rho > 0$ with $B(x^*, \rho) \subset U_{\alpha_0}$. Choose $j$ large enough that $\|x_{m_j} - x^*\| < \rho/2$ and $1/m_j < \rho/2$ simultaneously (possible since $x_{m_j}\to x^*$ and $m_j \to \infty$ along the subsequence). Then for any $y \in B(x_{m_j}, 1/m_j)$,

so $B(x_{m_j}, 1/m_j) \subset B(x^*,\rho) \subset U_{\alpha_0}$. Thus $B(x_{m_j}, 1/m_j) \cap K$ is covered by the single set $U_{\alpha_0}$ — a finite subcollection of $\mathcal{U}$ — contradicting the defining property of $x_{m_j}$ chosen above. This contradiction shows some finite subcollection of $\mathcal{U}$ must cover $K$ after all.

(c) $\Rightarrow$ (a). Assume $K$ has the open-cover finite-subcover property. Boundedness. The balls $U_k = B(0,k)$ for $k = 1, 2, 3, \dots$ form an open cover of $\mathbb{R}^n$ (every point has finite norm, hence lies in some $B(0,k)$), hence of $K$. By hypothesis, finitely many of them cover $K$: $K \subset B(0,k_1) \cup \cdots \cup B(0,k_m) = B(0,\max_i k_i)$. Hence $K$ is bounded. Closedness. Suppose $K$ is not closed; then there is a limit point $p \in \mathbb{R}^n \setminus K$ of $K$. For each $k \in \mathbb{N}$, let

an open set (complement of a closed ball). Since $p \notin K$, every $x \in K$ satisfies $x \ne p$, so $\|x-p\| > 0$, so $x \in U_k$ for $k$ large enough; hence $\{U_k\}$ is an open cover of $K$. If finitely many $U_{k_1}, \dots, U_{k_m}$ covered $K$, then with $k^* = \max_i k_i$ we'd have $K \subset U_{k^*} = \{x : \|x-p\| > 1/k^*\}$, meaning $K \cap B(p, 1/k^*) = \emptyset$. But $p$ is a limit point of $K$, so every ball around $p$ meets $K$ at a point other than $p$ — contradiction. So no finite subcover exists, contradicting hypothesis (c). Hence $K$ must be closed. $\blacksquare$

By the Heine–Borel theorem, $K$ is closed and bounded. Since $F \subset K$, $F$ is bounded. Since $F$ is closed by hypothesis, $F$ is closed and bounded, hence compact by Heine–Borel. $\blacksquare$