V\}\in \mathscr{N}(\mathscr{V})$, see [@dowker:47]. Then the composite map $f=h\circ g$ is a continuous $\varepsilon$-selection for $\varphi$. Indeed, $x\in \min \sigma_x$ and by and , we get that $f(x)=h(g(x))\in h(|\sigma_x|)\subset \Psi(\min \sigma_x)\subset
  \mathbf{O}_{\varepsilon(x)}(\varphi(x))$.

\[remark-Near-sel-vgg:2\] For an infinite cardinal $\tau$, we shall say that a subset $S$ of a metric space $(Y,\rho)$ is *uniformly $\tau$-$UV^\omega$* if for each $\varepsilon>0$ there exists $\delta\in(0,\varepsilon]$ such that for every finite-dimensional simplicial complex $\Sigma$ with $\operatorname{Card}(\Sigma^0)<\tau$, every continuous map $g:|\Sigma|\to \mathbf{O}_\delta(S)$ is null-homotopic in $\mathbf{O}_\varepsilon(S)$. Also, let us recall that the *degree of compactness* of space $X$ is the least infinite cardinal $\kappa(X)$ such that every open cover of $X$ has an open refinement of cardinality less than ${\kappa(X)}$, see [@choban-mihaylova-nedev:08]. Using the same proof, Theorem \[theorem-Finite-C-v5:1\] remains valid if each $\varphi(x)$, $x\in X$, is only assumed to be uniformly $\kappa(X)$-$UV^\omega$. Indeed, if $\mathscr{V}_n$, $n\leq k$, are as in that proof, then ${\operatorname{Card}(\mathscr{V})<\kappa(X)}$ because each family $\mathscr{V}_n$, $n\leq k$, is discrete. Hence, according to Remark \[remark-Near-sel-v10:1\], we can still apply Theorem \[theorem-Finite-C-v6:1\] to get a continuous map $h:|\mathscr{N}(\mathscr{V})|\to Y$ satisfying with respect to the oriented simplicial complex $\mathscr{N}(\mathscr{V})$. Having already the map $h$, we can finish the proof precisely as before. One benefit of this refined 