[proofplan] The forward direction follows from two facts: compact subsets of Hausdorff spaces are closed, and a finite subcover of $\{B(0, m)\}$ gives boundedness. The reverse direction proceeds in three stages: prove $[a,b]$ is compact by the supremum (least upper bound) argument, apply [Tychonoff for finite products](/theorems/312) inductively to get compactness of $[-M, M]^n$, and invoke the [closed-subset-of-compact theorem](/theorems/307) to conclude. [/proofplan] [step:Show compact implies closed and bounded] **Closed:** $\mathbb{R}^n$ is Hausdorff (the Euclidean metric induces the standard [topology](/page/Topology)). By [compact subspaces and Hausdorff spaces](/theorems/307) (part 2), compact subsets of Hausdorff spaces are closed. **Bounded:** The nested open cover $\{B(0, m)\}_{m \in \mathbb{N}}$ covers $K$. Compactness gives a finite subcover: $K \subseteq B(0, m_1) \cup \cdots \cup B(0, m_k) = B(0, M)$ where $M = \max(m_1, \ldots, m_k)$. [/step] [step:Prove $[a, b]$ is compact using 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 $S = \{x \in [a, b] : [a, x] \text{ has a finite subcover from } \mathcal{U}\}$. Since some $U \in \mathcal{U}$ contains $a$, we have $a \in S$, so $S \neq \varnothing$. Let $s = \sup S$. Since $s \in [a, b]$, choose $U_0 \in \mathcal{U}$ with $s \in U_0$, and $\varepsilon > 0$ with $(s - \varepsilon, s + \varepsilon) \cap [a, b] \subseteq U_0$. By the definition of supremum, there exists $x_0 \in S$ with $x_0 > s - \varepsilon$. A finite subcover $\mathcal{F}$ of $[a, x_0]$ exists; then $\mathcal{F} \cup \{U_0\}$ covers $[a, \min(s + \varepsilon, b)]$. If $s < b$, then $\min(s + \varepsilon, b) \in S$ with $\min(s + \varepsilon, b) > s$, contradicting $s = \sup S$. So $s = b$ and $b \in S$, meaning $[a, b]$ has a finite subcover. [/proof] [/step] [step:Deduce compactness of $[-M, M]^n$ via Tychonoff for finite products] The interval $[-M, M]$ is compact (previous step). By [Tychonoff's theorem for finite products](/theorems/312), the product of two compact spaces is compact. Applying inductively: $[-M, M]^n$ is compact. [/step] [step:Conclude $K$ is compact as a closed subset of $[-M, M]^n$] Since $K$ is bounded, $K \subseteq [-M, M]^n$ for some $M > 0$. The set $K$ is closed in $\mathbb{R}^n$ by hypothesis, hence closed in $[-M, M]^n$. By [compact subspaces and Hausdorff spaces](/theorems/307) (part 1), a closed subset of a [compact space](/page/Compact%20Space) is compact. Therefore $K$ is compact. [/step]