[proofplan] The forward direction uses that $\mathbb{R}^n$ is Hausdorff (so compact subsets are closed) and that the open cover $\{B(0, m)\}$ extracts a finite subcover (giving boundedness). The reverse direction builds in three stages: prove $[a,b]$ is compact using a supremum argument, apply [Tychonoff for finite products](/theorems/310) inductively to get compactness of $[-M,M]^n$, then use the [closed-subset-of-compact theorem](/theorems/307) to conclude $K$ is compact. [/proofplan] [step:Show compact implies closed and bounded] **Closed:** $\mathbb{R}^n$ is Hausdorff (the metric $d(x,y) = \|x - y\|$ induces the standard topology). By [compact subspaces and Hausdorff spaces](/theorems/307) (part 2), compact subsets of Hausdorff spaces are closed. Therefore $K$ is closed. **Bounded:** The collection $\{B(0, m)\}_{m \in \mathbb{N}}$ is an open cover of $K$ (for each $x \in K$, $\|x\| < m$ for large $m$). By compactness, finitely many suffice: $K \subseteq B(0, m_1) \cup \cdots \cup B(0, m_k)$. Since the balls are nested, $K \subseteq B(0, M)$ where $M = \max(m_1, \ldots, m_k)$. So $K$ is bounded. [/step] [step:Prove $[a, b]$ is compact via the supremum argument] [claim:Compactness of $[a, b]$] Every closed bounded interval $[a, b] \subseteq \mathbb{R}$ is compact. [/claim] [proof] Let $\mathcal{U}$ be an open cover of $[a, b]$. Define \begin{align*} S = \{x \in [a, b] : [a, x] \text{ can be covered by finitely many sets from } \mathcal{U}\}. \end{align*} The set $S$ is non-empty: since $a \in [a,b]$, some $U_0 \in \mathcal{U}$ contains $a$, and $\{U_0\}$ covers $[a, a]$, so $a \in S$. Let $c = \sup S$. We show $c = b$ and $c \in S$. Since $c \in [a, b]$, there exists $U_c \in \mathcal{U}$ with $c \in U_c$. Since $U_c$ is open, $(c - \varepsilon, c + \varepsilon) \cap [a, b] \subseteq U_c$ for some $\varepsilon > 0$. By definition of the supremum, there exists $s \in S$ with $s > c - \varepsilon$. Then $[a, s]$ has a finite subcover $\mathcal{F} \subseteq \mathcal{U}$, and $\mathcal{F} \cup \{U_c\}$ covers $[a, \min(c + \varepsilon, b)]$. If $c < b$, then $\min(c + \varepsilon, b) > c$ belongs to $S$ (possibly after shrinking $\varepsilon$), contradicting $c = \sup S$. Therefore $c = b$. Moreover, $b \in S$ because $\mathcal{F} \cup \{U_c\}$ covers $[a, b]$. [/proof] [guided] The supremum argument is an alternative to the bisection proof. We define $S = \{x \in [a, b] : [a, x] \text{ can be covered by finitely many sets from } \mathcal{U}\}$ and let $c = \sup S$. **$S$ is non-empty:** Since $a \in [a, b]$, some $U_0 \in \mathcal{U}$ contains $a$, and $\{U_0\}$ covers $[a, a]$. So $a \in S$. **$c = b$ (cannot stop early):** Suppose $c < b$. Since $c \in [a, b]$, some $U_c \in \mathcal{U}$ contains $c$, and $U_c$ contains an interval $(c - \varepsilon, c + \varepsilon) \cap [a, b]$. By definition of $\sup$, there exists $s \in S$ with $s > c - \varepsilon$. The initial segment $[a, s]$ has a finite subcover $\mathcal{F}$, and $\mathcal{F} \cup \{U_c\}$ covers $[a, \min(c + \varepsilon, b)]$. This places $\min(c + \varepsilon, b) \in S$, contradicting $c = \sup S$ if $c < b$. **$b \in S$:** Once $c = b$, the same argument shows $b \in S$: some $U_b \in \mathcal{U}$ covers $b$ and extends slightly to the left, and combining with a finite cover of $[a, s]$ (for $s$ close to $b$) produces a finite cover of all of $[a, b]$. Therefore $[a, b]$ has a finite subcover, proving compactness. [/guided] [/step] [step:Deduce $[-M, M]^n$ is compact by Tychonoff for finite products] By the previous step, $[-M, M]$ is compact. By [Tychonoff's theorem for finite products](/theorems/310), the product of two compact spaces is compact. Applying this inductively: $[-M, M]^2 = [-M, M] \times [-M, M]$ is compact, $[-M, M]^3 = [-M, M]^2 \times [-M, M]$ is compact, and so on. After $n - 1$ applications, $[-M, M]^n$ is compact. [/step] [step:Conclude $K$ is compact as a closed subset of the compact box $[-M, M]^n$] Since $K$ is bounded, $K \subseteq [-M, M]^n$ for some $M > 0$. The box $[-M, M]^n$ is compact (previous step). The set $K$ is closed in $\mathbb{R}^n$ by hypothesis, and since $[-M, M]^n$ is a subspace of $\mathbb{R}^n$, $K$ is also closed in $[-M, M]^n$ (a set closed in the ambient space is closed in any subspace containing it). By [compact subspaces and Hausdorff spaces](/theorems/307) (part 1), a closed subset of a compact space is compact. Therefore $K$ is compact. [/step]