[proofplan] We prove each property separately. Monotonicity follows directly from the monotonicity of $\mathcal{H}^s$. Countable stability combines the monotonicity of dimension (giving $\geq$) with countable subadditivity of $\mathcal{H}^s$ (giving $\leq$). Countable sets have dimension zero because single points have $\mathcal{H}^s(\{x\}) = 0$ for all $s > 0$, and countable stability applies. Open sets contain balls, which have $\mathcal{H}^n$ equal to $\mathcal{L}^n > 0$ by [$\mathcal{H}^n = \mathcal{L}^n$ on $\mathbb{R}^n$](/theorems/3048), giving $\dim_{\mathcal{H}} \geq n$, and the bound $\dim_{\mathcal{H}} \leq n$ follows from a covering argument. Smooth $k$-dimensional manifolds are locally bi-Lipschitz to open subsets of $\mathbb{R}^k$, so [Bi-Lipschitz Maps Preserve Hausdorff Dimension](/theorems/3061) and countable stability give $\dim_{\mathcal{H}}(M) = k$. [/proofplan] [step:Prove monotonicity: $A \subset B$ implies $\dim_{\mathcal{H}}(A) \leq \dim_{\mathcal{H}}(B)$] Let $t > \dim_{\mathcal{H}}(B)$. By definition of Hausdorff dimension, $\mathcal{H}^t(B) = 0$. Since $A \subset B$, monotonicity of $\mathcal{H}^t$ (established in [$\mathcal{H}^s$ is a Borel Regular Outer Measure](/theorems/3044)) gives $\mathcal{H}^t(A) \leq \mathcal{H}^t(B) = 0$, so $\mathcal{H}^t(A) = 0$. Since this holds for every $t > \dim_{\mathcal{H}}(B)$, the definition $\dim_{\mathcal{H}}(A) = \inf\{t \geq 0 : \mathcal{H}^t(A) = 0\}$ gives $\dim_{\mathcal{H}}(A) \leq \dim_{\mathcal{H}}(B)$. [/step] [step:Prove countable stability: $\dim_{\mathcal{H}}\!\left(\bigcup_{j=1}^\infty E_j\right) = \sup_j \dim_{\mathcal{H}}(E_j)$] Write $E = \bigcup_{j=1}^\infty E_j$ and $d = \sup_{j \geq 1} \dim_{\mathcal{H}}(E_j)$. **Lower bound** ($\dim_{\mathcal{H}}(E) \geq d$): Each $E_j \subset E$, so by monotonicity, $\dim_{\mathcal{H}}(E_j) \leq \dim_{\mathcal{H}}(E)$ for all $j$. Taking the supremum over $j$ gives $d \leq \dim_{\mathcal{H}}(E)$. **Upper bound** ($\dim_{\mathcal{H}}(E) \leq d$): Let $t > d$. Then $t > \dim_{\mathcal{H}}(E_j)$ for every $j$, so $\mathcal{H}^t(E_j) = 0$ for all $j$. By countable subadditivity of $\mathcal{H}^t$ (from [$\mathcal{H}^s$ is a Borel Regular Outer Measure](/theorems/3044)): \begin{align*} \mathcal{H}^t(E) = \mathcal{H}^t\!\left(\bigcup_{j=1}^\infty E_j\right) \leq \sum_{j=1}^\infty \mathcal{H}^t(E_j) = 0. \end{align*} Since $\mathcal{H}^t(E) = 0$ for all $t > d$, we have $\dim_{\mathcal{H}}(E) \leq d$. [/step] [step:Prove that countable sets have dimension zero] Let $E = \{x_j\}_{j=1}^\infty$ be a countable subset of $\mathbb{R}^n$ (the case of finite $E$ is included by allowing eventually constant sequences). It suffices to show $\dim_{\mathcal{H}}(\{x\}) = 0$ for every single point $x \in \mathbb{R}^n$, and then apply countable stability. Fix $x \in \mathbb{R}^n$ and $s > 0$. For any $\delta > 0$, the singleton cover $\{C_1\} = \{\{x\}\}$ satisfies $\operatorname{diam}(\{x\}) = 0 \leq \delta$, so it is a valid $\delta$-cover of $\{x\}$. The covering sum is $\alpha(s)(0/2)^s = 0$. Therefore $\mathcal{H}^s_\delta(\{x\}) = 0$ for all $\delta > 0$, giving $\mathcal{H}^s(\{x\}) = 0$. Since $\mathcal{H}^s(\{x\}) = 0$ for all $s > 0$, we have $\dim_{\mathcal{H}}(\{x\}) = \inf\{s \geq 0 : \mathcal{H}^s(\{x\}) = 0\} = 0$. By countable stability: $\dim_{\mathcal{H}}(E) = \sup_j \dim_{\mathcal{H}}(\{x_j\}) = \sup_j 0 = 0$. [/step] [step:Prove that non-empty open sets have dimension $n$] Let $U \subset \mathbb{R}^n$ be open and non-empty. Since $U$ is open, there exists a point $x_0 \in U$ and a radius $r_0 > 0$ such that the closed ball $\overline{B}(x_0, r_0) \subset U$. **Lower bound** ($\dim_{\mathcal{H}}(U) \geq n$): By monotonicity, $\dim_{\mathcal{H}}(U) \geq \dim_{\mathcal{H}}(\overline{B}(x_0, r_0))$. By [$\mathcal{H}^n = \mathcal{L}^n$ on $\mathbb{R}^n$](/theorems/3048), $\mathcal{H}^n(\overline{B}(x_0, r_0)) = \mathcal{L}^n(\overline{B}(x_0, r_0)) = \alpha(n) r_0^n > 0$. Since $\mathcal{H}^n(\overline{B}(x_0, r_0)) > 0$, the [jump theorem](/theorems/3045) (applied contrapositively) gives $\dim_{\mathcal{H}}(\overline{B}(x_0, r_0)) \geq n$. Therefore $\dim_{\mathcal{H}}(U) \geq n$. **Upper bound** ($\dim_{\mathcal{H}}(U) \leq n$): Fix $t > n$. We show $\mathcal{H}^t(U) = 0$. Write $U = \bigcup_{m=1}^\infty K_m$ where $K_m = \{x \in U : |x| \leq m, \operatorname{dist}(x, \partial U) \geq 1/m\}$ is compact. It suffices to show $\mathcal{H}^t(K_m) = 0$ for each $m$, since countable subadditivity then gives $\mathcal{H}^t(U) = 0$. Each $K_m$ is bounded: $K_m \subset \overline{B}(0, m)$. By [$\mathcal{H}^n = \mathcal{L}^n$ on $\mathbb{R}^n$](/theorems/3048), $\mathcal{H}^n(K_m) = \mathcal{L}^n(K_m) \leq \mathcal{L}^n(\overline{B}(0, m)) = \alpha(n) m^n < +\infty$. Since $\mathcal{H}^n(K_m) < +\infty$ and $t > n$, the [jump theorem](/theorems/3045) gives $\mathcal{H}^t(K_m) = 0$. Therefore $\dim_{\mathcal{H}}(U) \leq n$, and combining both bounds: $\dim_{\mathcal{H}}(U) = n$. [/step] [step:Prove that smooth $k$-dimensional embedded submanifolds have dimension $k$] Let $M \subset \mathbb{R}^n$ be a smooth $k$-dimensional embedded submanifold (without boundary). By the definition of an embedded submanifold, for every $p \in M$ there exist a neighborhood $V_p \subset M$ of $p$ (in the subspace topology) and a diffeomorphism $\varphi_p : V_p \to W_p$ onto an open set $W_p \subset \mathbb{R}^k$. Since $M$ is second-countable (as a subspace of $\mathbb{R}^n$), we may extract a countable subcover: there exist countably many points $\{p_j\}_{j=1}^\infty$ such that $M = \bigcup_{j=1}^\infty V_{p_j}$. By countable stability, $\dim_{\mathcal{H}}(M) = \sup_j \dim_{\mathcal{H}}(V_{p_j})$. We show $\dim_{\mathcal{H}}(V_{p_j}) = k$ for each $j$. Fix $j$ and write $V = V_{p_j}$, $W = W_{p_j}$, $\varphi = \varphi_{p_j}$. Since $\varphi: V \to W$ is a diffeomorphism between a subset of $\mathbb{R}^n$ and an open subset $W$ of $\mathbb{R}^k$, and $V$ is compact in no neighborhood, we work locally. Choose an open set $W' \subset\subset W$ (compactly contained in $W$) and let $V' = \varphi^{-1}(W')$. On $\overline{W'}$, the map $\varphi^{-1}: \overline{W'} \to \mathbb{R}^n$ is smooth with bounded derivative, hence Lipschitz with some constant $L_1 > 0$. Similarly, $\varphi: \overline{V'} \to \mathbb{R}^k$ is Lipschitz with some constant $L_2 > 0$. Therefore $\varphi|_{V'}: V' \to W'$ is bi-Lipschitz with constants $1/L_1 \leq L_2$ (after possibly relabeling). By [Bi-Lipschitz Maps Preserve Hausdorff Dimension](/theorems/3061), $\dim_{\mathcal{H}}(V') = \dim_{\mathcal{H}}(W')$. Since $W' \subset \mathbb{R}^k$ is open and non-empty, the previous property gives $\dim_{\mathcal{H}}(W') = k$, so $\dim_{\mathcal{H}}(V') = k$. Since $V = \bigcup_{m=1}^\infty V'_m$ where $V'_m = \varphi^{-1}(W'_m)$ for an exhaustion $W'_1 \subset W'_2 \subset \cdots$ of $W$ by compactly contained open sets, countable stability gives $\dim_{\mathcal{H}}(V) = \sup_m \dim_{\mathcal{H}}(V'_m) = k$. Therefore $\dim_{\mathcal{H}}(M) = \sup_j k = k$. [/step]