\bigcap_{j \in J} A_j.$$ An application of Lemma \[lem:deformation\] finishes the proof.

\[lem:uclose uhomotopic\] If ${{\mathcal{U}_{}}}$ is an open $AE(n)$-cover of a space $Y$, then any two ${{\mathcal{U}_{}}}$-close maps from a metric space of dimension less than $n$ into $Y$ are ${{\mathcal{U}_{}}}$-homotopic.

Let ${{\mathcal{U}_{}}}$ be an open $AE(n)$-cover of a metric space $Y$. Let $X$ be a metric space of dimension less than $n$. Let $f, g \colon X \to Y$ be ${{\mathcal{U}_{}}}$-close. We have to show that $f$ and $g$ are homotopic by a homotopy whose paths refine ${{\mathcal{U}_{}}}$.

Let ${{\mathcal{V}_{}}} = \{ V_U \}_{U \in {{\mathcal{U}_{}}}}$ be the collection of subsets of $X$ defined by $V_U = f^{-1}(U) \cap g^{-1}(U)$. Since $f$ and $g$ are ${{\mathcal{U}_{}}}$-close, ${{\mathcal{V}_{}}}$ is an open cover of $X$.

Let ${{\mathcal{W}_{}}}$ be a closed, locally finite cover of $X$ with multiplicity at most $n$ that refines ${{\mathcal{V}_{}}}$. Let ${{\mathcal{F}_{}}} = \{ W \times [0,1] \colon W \in {{\mathcal{W}_{}}} \}$ be a cover of $X \times [0,1]$. Note that ${{\mathcal{F}_{}}}$ is a closed, locally finite, locally finite-dimensional cover of $X \times [0,1]$.

Let $H_0 \colon K \times \{ 0, 1 \} \to Y$ be the map defined by $H_0(x, 0) = f(x)$ and $H_0(x,1) = 