Why using closure of U to define the topology on the orientation bundle of a topological manifold? Sorry I have to quote a major part of content from Bredon’s /Topology and Geometry/ as a screenshot. In the following definition of the basis $U_\alpha$, why using the closure of $U$ but not $U$ directly? It seems the arguments for $U_\alpha$ being a basis still hold with $j_{x, U}$.
Thanks!

edit: I added another screenshot on the proof of $U_\alpha$ being a basis. It may require taking the closure of $U$ somewhere. But I cannot find it.

 A: Your are right.
For all $R \subset S \subset M$ let us denote by $j_{R,S} : H_n(M,M-S;G) \to H_n(M,M-R;G)$ the map induced from inclusion. Then $j_{x,A} = j_{\{x\},A}$ (this notation can be used also for non-closed $A$). Let $j_U = j_{U, \overline U} : H_n(M,M-\overline U;G) \to H_n(M,M- U;G)$.
Let us define $U'_\beta = \{j_{x,U}(\beta) \mid x \in U \}$ for all $\beta \in H_n(M,M- U;G)$. For $\alpha \in H_n(M,M-\overline U;G)$ we then have
$$U_\alpha = U'_{j_U(\alpha)} .$$
Thus the set $\mathcal U = \{U_\alpha \mid U \subset M \text{ open, } \alpha \in H_n(M,M-\overline U;G)\}$ is a subset of $\mathcal U' = \{U'_\beta \mid  U \subset M \text{ open, } \beta \in H_n(M,M- U;G)\}$.
Bredon has proved in Proposition 7.3 that $\mathcal U$ is the basis for a topology on $\Theta \otimes G$. We shall prove (by modifying Bredon's arguments) that also  $\mathcal U'$ is the basis for a topology on $\Theta \otimes G$ and the topologies generated by $\mathcal U, \mathcal U'$ agree.
To do so, we shall show that for any two $U'_\alpha, V'_\beta \in \mathcal U'$ and each $\tilde x \in U'_\alpha \cap V'_\beta$ we can find an open neigborhood $W$ of $x$ and $\gamma \in H_n(M,M-\overline W;G)$ such that $\tilde x  \in W_\gamma \subset  U'_\alpha \cap V'_\beta$.
Since $W_\gamma \in \mathcal U \subset \mathcal U'$, we conclude that $\mathcal U'$ forms a basis. Moreover, taking $U =  V$ and $\alpha = \beta$, we see that each $U'_\alpha$ is the union of suitable $W_\gamma$. This shows the topologies generated by $\mathcal U, \mathcal U'$ agree.
So let us consider $\tilde x \in U'_\alpha \cap V'_\beta$. The element $\tilde x$ lies in a unique $\Theta_x \otimes G$. We thus have $\tilde x = j_{x,U}(\alpha) = j_{x,V}(\beta)$. We can find a convex open set $W$ in a euclidean neigborhood of $x$ in $M$ such that $\overline W \subset U \cap V$. Then $j_{x,\overline W}$ is an isomorphism. Let $\gamma \in H_n(M,M - \overline W;G)$ be the unique element such that $j_{x,\overline W}(\gamma) = \tilde x$. In particular $\tilde x \in W_\gamma$.
The homomorphism $j_{\overline W,U} : H_n(M,M-U) \to H_n(M,M-\overline W)$ has the property $j_{\overline W,U}(\alpha) = \gamma$ because $j_{x,\overline W}(\gamma) = \tilde x = j_{x,U} (\alpha) = j_{x,\overline W}(j_{\overline W,U}(\alpha))$. Now let $\tilde y \in W_\gamma$. We have $\tilde y = j_{y,\overline W}(\gamma)$ for a unique $y \in W$. Thus $\tilde y = j_{y,\overline W}(j_{\overline W,U}(\alpha)) = j_{y,U}(\alpha) \in U'_\alpha$. Similarly $\tilde y \in V'_\beta$.
