[proofplan] We construct the partition of unity in three stages. First, using $\sigma$-compactness, we build an exhaustion $\{K_j\}$ by the [Exhaustion by Compact Sets](/theorems/1062) theorem and refine the given cover to a locally finite open cover indexed by the "annular" regions $K_{j+1} \setminus \operatorname{int}(K_{j-1})$. Second, we apply [Urysohn's Lemma for Locally Compact Hausdorff Spaces](/theorems/1065) on each annular region to produce compactly supported bump functions. Third, we normalise by the (locally finite, hence everywhere positive) sum to obtain the partition of unity. [/proofplan] [step:Build an exhaustion and a locally finite refinement of the cover] By the [Exhaustion by Compact Sets](/theorems/1062) theorem, there exists a sequence of compact sets $\{K_j\}_{j=1}^\infty$ with $K_j \subset \operatorname{int}(K_{j+1})$ for all $j$ and $X = \bigcup_{j=1}^\infty K_j$. Set $K_0 := \varnothing$ and $K_{-1} := \varnothing$ for notational convenience. For each $j \ge 1$, define the open "annular" region \begin{align*} A_j := \operatorname{int}(K_{j+1}) \setminus K_{j-1}. \end{align*} Each $A_j$ is open in $X$ (it is the intersection of the open set $\operatorname{int}(K_{j+1})$ and the open set $X \setminus K_{j-1}$, where $K_{j-1}$ is compact in the Hausdorff space $X$, hence closed by [Compact Subspaces and Hausdorff Spaces](/theorems/307)). The compact "strip" $K_j \setminus \operatorname{int}(K_{j-1})$ is contained in $A_j$, and $\bigcup_{j=1}^\infty (K_j \setminus \operatorname{int}(K_{j-1})) = X$ (since every $x \in X$ lies in $K_j$ for some minimal $j$). Therefore $\{A_j\}_{j=1}^\infty$ covers $X$. For each $x \in K_j \setminus \operatorname{int}(K_{j-1})$, the point $x$ lies in $A_j$ and in some $U_\alpha$. Choose an open set $V_x \subset A_j \cap U_\alpha$ containing $x$ such that $\overline{V_x}$ is compact and $\overline{V_x} \subset A_j \cap U_\alpha$. Such $V_x$ exists by the [Urysohn's Lemma for Locally Compact Hausdorff Spaces](/theorems/1065) construction (specifically, the interposition step): the locally compact Hausdorff space $X$ allows us to find a precompact open neighbourhood of $x$ inside any open set. The compact set $K_j \setminus \operatorname{int}(K_{j-1})$ is covered by finitely many such $V_x$; call them $V_{j,1}, \ldots, V_{j,m_j}$, each with an associated index $\alpha(j,k) \in A$ satisfying $\overline{V_{j,k}} \subset A_j \cap U_{\alpha(j,k)}$. The collection $\{V_{j,k}\}_{j \ge 1, 1 \le k \le m_j}$ is an open cover of $X$ that refines $\{U_\alpha\}$. It is **locally finite**: any point $x \in X$ lies in some $K_N$ (by the exhaustion), hence $x \in \operatorname{int}(K_{N+1})$. The open set $\operatorname{int}(K_{N+1})$ can intersect $A_j = \operatorname{int}(K_{j+1}) \setminus K_{j-1}$ only when $j - 1 \le N + 1$, i.e., $j \le N + 2$. So $\operatorname{int}(K_{N+1})$ meets only finitely many of the $A_j$'s, and each $A_j$ contributes only finitely many $V_{j,k}$'s. [guided] The exhaustion $\{K_j\}$ provides a filtration of $X$ into compact "shells." The annular regions $A_j = \operatorname{int}(K_{j+1}) \setminus K_{j-1}$ overlap by design — each point of $X$ lies in at most three consecutive $A_j$'s — which is what makes the resulting cover locally finite. Why do we use $K_{j+1}$ and $K_{j-1}$ rather than $K_j$ and $K_{j-1}$? Because we need $K_j \setminus \operatorname{int}(K_{j-1}) \subset A_j$, and this requires $K_j \subset \operatorname{int}(K_{j+1})$ (from the exhaustion property) and $K_{j-1}$ to be removed. The "gap" between $K_{j-1}$ and $K_{j+1}$ ensures openness and provides room for the compact closures of the $V_{j,k}$. Local finiteness follows from the bounded overlap: the neighbourhood $\operatorname{int}(K_{N+1})$ of any point in $K_N$ meets $A_j$ only for $j \le N+2$, and each $A_j$ contributes finitely many open sets. [/guided] [/step] [step:Construct compactly supported bump functions for each element of the refinement] For each $(j, k)$, apply the [Urysohn's Lemma for Locally Compact Hausdorff Spaces](/theorems/1065) with the compact set $\overline{V_{j,k}}$ and the open set $A_j \cap U_{\alpha(j,k)}$ (note that $\overline{V_{j,k}} \subset A_j \cap U_{\alpha(j,k)}$ by construction). This yields a continuous function $\phi_{j,k}: X \to [0,1]$ with $\phi_{j,k} \equiv 1$ on $\overline{V_{j,k}}$, $\operatorname{supp}(\phi_{j,k}) \subset A_j \cap U_{\alpha(j,k)}$, and $\operatorname{supp}(\phi_{j,k})$ compact. Define the sum \begin{align*} \Phi(x) := \sum_{j=1}^\infty \sum_{k=1}^{m_j} \phi_{j,k}(x). \end{align*} By local finiteness of the supports, at each point $x$ only finitely many terms are nonzero, so $\Phi$ is well-defined and continuous. Moreover, $\Phi(x) \ge 1$ for all $x \in X$: for any $x$, there exist $j$ and $k$ with $x \in V_{j,k}$ (since $\{V_{j,k}\}$ covers $X$), and $\phi_{j,k}(x) = 1$ on $\overline{V_{j,k}} \supset V_{j,k} \ni x$. Therefore $\Phi(x) \ge \phi_{j,k}(x) = 1 > 0$. [/step] [step:Normalise and reindex to produce the partition of unity] For each $\alpha \in A$, define \begin{align*} \psi_\alpha(x) := \frac{1}{\Phi(x)} \sum_{\substack{(j,k) : \alpha(j,k) = \alpha}} \phi_{j,k}(x). \end{align*} Since $\Phi(x) > 0$ for all $x$ and the sum in the numerator has only finitely many nonzero terms at each $x$ (by local finiteness), $\psi_\alpha$ is well-defined and continuous. **Values in $[0,1]$.** Each $\phi_{j,k}(x) \ge 0$ and $\Phi(x) \ge 1$, so $\psi_\alpha(x) \ge 0$. Also, $\sum_{\alpha \in A} \psi_\alpha(x) = \frac{1}{\Phi(x)} \sum_{j,k} \phi_{j,k}(x) = \frac{\Phi(x)}{\Phi(x)} = 1$, so each $\psi_\alpha(x) \le 1$. **Compact support in $U_\alpha$.** For each $(j,k)$ with $\alpha(j,k) = \alpha$, we have $\operatorname{supp}(\phi_{j,k}) \subset U_\alpha$. The support of $\psi_\alpha$ is contained in $\overline{\bigcup_{(j,k):\alpha(j,k)=\alpha} \operatorname{supp}(\phi_{j,k})}$. By local finiteness, near any point only finitely many supports are present, so the closure of the union is the union of the closures: $\operatorname{supp}(\psi_\alpha) \subset \bigcup_{(j,k):\alpha(j,k)=\alpha} \operatorname{supp}(\phi_{j,k}) \subset U_\alpha$. Each $\operatorname{supp}(\phi_{j,k})$ is compact and $\operatorname{supp}(\psi_\alpha)$ is a closed subset of the locally finite union, so it is closed. Since the supports cluster only in finitely many annuli near any compact set, $\operatorname{supp}(\psi_\alpha)$ is compact (it is a closed subset of a finite union of compact sets for any given compact region). **Local finiteness.** The collection $\{\operatorname{supp}(\psi_\alpha)\}_{\alpha \in A}$ is locally finite because $\{\operatorname{supp}(\phi_{j,k})\}$ is locally finite and each $\operatorname{supp}(\psi_\alpha)$ is a union of sets from this locally finite family with the same index $\alpha$. **Partition of unity.** As computed above, $\sum_{\alpha \in A} \psi_\alpha(x) = 1$ for all $x \in X$. [/step]