[proofplan] The forward direction uses two standard facts: compact subsets of Hausdorff spaces are closed, and compact sets admit finite subcovers from the cover by balls $\{B(0, m)\}$, giving boundedness. The reverse direction proceeds in three stages: first, prove $[a,b]$ is compact by the bisection (nested intervals) argument; second, apply [Tychonoff's theorem for finite products](/theorems/308) to deduce $[-M, M]^n$ is compact; third, invoke the [closed-subset-of-compact theorem](/theorems/307) to conclude $K$ is compact. [/proofplan] [step:Show compact implies closed using the Hausdorff property of $\mathbb{R}^n$] The space $\mathbb{R}^n$ is Hausdorff (it is metrisable). By [compact subspaces and Hausdorff spaces](/theorems/307) (part 2), compact subsets of Hausdorff spaces are closed. Therefore $K$ is closed. [/step] [step:Show compact implies bounded by extracting a finite subcover from the ball cover] The collection $\{B(0, m)\}_{m \in \mathbb{N}}$ is an open cover of $K$ (for each $x \in K$, $\|x\| < m$ for sufficiently large $m$). By compactness, finitely many suffice: $K \subseteq B(0, m_1) \cup \cdots \cup B(0, m_k) = B(0, M)$ where $M = \max(m_1, \ldots, m_k)$. Therefore $K$ is bounded. [/step] [step:Prove $[a,b]$ is compact by the bisection argument] [claim:Compactness of $[a,b]$] Every closed bounded interval $[a, b] \subseteq \mathbb{R}$ is compact. [/claim] [proof] Suppose for contradiction that $\mathcal{U}$ is an open cover of $[a, b]$ with no finite subcover. Bisect $[a, b]$ into $[a, (a+b)/2]$ and $[(a+b)/2, b]$. At least one half has no finite subcover from $\mathcal{U}$; call it $[a_1, b_1]$. Iterate: at stage $n$, bisect $[a_{n-1}, b_{n-1}]$ and choose the half $[a_n, b_n]$ with no finite subcover. The nested intervals satisfy $b_n - a_n = (b - a)/2^n \to 0$. By the nested intervals theorem, $\bigcap_{n=0}^{\infty} [a_n, b_n] = \{c\}$ for some $c \in [a, b]$. Since $\mathcal{U}$ covers $[a, b]$, there exists $U \in \mathcal{U}$ with $c \in U$. Since $U$ is open in $\mathbb{R}$, there exists $\varepsilon > 0$ with $(c - \varepsilon, c + \varepsilon) \subseteq U$. For $n$ large enough that $(b - a)/2^n < \varepsilon$, we have $[a_n, b_n] \subseteq (c - \varepsilon, c + \varepsilon) \subseteq U$, so $\{U\}$ is a finite subcover of $[a_n, b_n]$. This contradicts the choice of $[a_n, b_n]$ as having no finite subcover. [/proof] [guided] The bisection argument is a proof by contradiction that uses the completeness of $\mathbb{R}$ (via the nested intervals theorem). **Setup:** Suppose for contradiction that $\mathcal{U}$ is an open cover of $[a, b]$ with no finite subcover. Bisect $[a, b]$ into two halves. At least one half has no finite subcover (if both did, their union would be a finite subcover of $[a, b]$). Call it $[a_1, b_1]$ and iterate. **Nested intervals:** At stage $n$, we have $[a_n, b_n]$ with no finite subcover and $b_n - a_n = (b - a)/2^n$. The sequences $(a_n)$ and $(b_n)$ are monotone and bounded, hence convergent by the completeness of $\mathbb{R}$. Their common limit $c = \lim a_n = \lim b_n$ is the unique point in $\bigcap_{n=0}^\infty [a_n, b_n]$. **Contradiction:** Since $\mathcal{U}$ covers $[a, b]$, some $U \in \mathcal{U}$ contains $c$. Since $U$ is open, there exists $\varepsilon > 0$ with $(c - \varepsilon, c + \varepsilon) \subseteq U$. For $n$ large enough that $(b - a)/2^n < \varepsilon$, the entire interval $[a_n, b_n]$ fits inside $(c - \varepsilon, c + \varepsilon) \subseteq U$. So $\{U\}$ is a finite subcover of $[a_n, b_n]$, contradicting the choice of $[a_n, b_n]$ as having no finite subcover. The nested intervals theorem requires completeness, which is why this argument works in $\mathbb{R}$ but not in $\mathbb{Q}$. [/guided] [/step] [step:Deduce $[-M, M]^n$ is compact via Tychonoff for finite products] By the previous step, $[-M, M]$ is compact. By [Tychonoff's theorem for finite products](/theorems/308), the product of two compact spaces is compact. Applying this inductively $n - 1$ times: \begin{align*} [-M, M]^n = [-M, M] \times [-M, M] \times \cdots \times [-M, M] \end{align*} is compact. [/step] [step:Conclude $K$ is compact as a closed subset of the compact space $[-M, M]^n$] Since $K$ is bounded, $K \subseteq [-M, M]^n$ for some $M > 0$. The space $[-M, M]^n$ is compact (by the previous step). Since $K$ is closed in $\mathbb{R}^n$, it is also closed in $[-M, M]^n$ (the subspace topology on $[-M, M]^n$ is finer on intersections with closed sets of the ambient space). By [compact subspaces and Hausdorff spaces](/theorems/307) (part 1), a closed subset of a compact space is compact. Therefore $K$ is compact. [/step]