I am looking in a book titled $\textit{Aspects of Topology}$ by Charles O. Christenson and William L. Voxman. In Chapter 6, they provide definitions and some properties of inverse limit spaces. Theorem 6.B.8 has me confused:
"$\textbf{(6.B.8) Theorem.}$ Suppose that $(X_\alpha, f_{\alpha\beta}, D)$ is an inverse system and that $X_\infty$ is the inverse limit. Then $\mathbf{B} = \{\hat{p}^{-1}_\alpha(U) \mid \alpha \in D \ \text{and} \ U \subset X_\alpha\}$ is a basis for $X_\infty$."
Here, $D$ is a directed set and $f_{\alpha\beta}: X_\beta \rightarrow X_\alpha$ where $\alpha \preceq \beta$ is a bonding map. $\hat{p}_\alpha: X_\infty \rightarrow X_\alpha$ is meant to be the projection map restricted to the inverse limit $X_\infty$.
I have a couple of issues with this theorem.
1---It seems like they meant to include that $U$ is an open subset of $X_\alpha$ in the definition of $\mathbf{B}$; however, if this was the case, then the theorem would certainly not be true in general (consider $X_\alpha = X$ for every $\alpha \in D$ and all bonding maps to be identity functions). It would make it a subbasis perhaps, but not a basis. This seems to be the case even without the assumption that $U$ is open.
2---There appears to be an obvious error on the fourth line from the bottom of the proof, where they say "$\displaystyle \bigcap_{i=1}^n \hat{p}^{-1}_\alpha f_{\alpha_i\alpha}^{-1}(U_{\alpha_i}) = \displaystyle \bigcap_{i=1}^n \hat{p}^{-1}_{\alpha_i}(U_{\alpha_i})$" and I speculate that this caused an error in the proof.
Can anyone either tell me either if I'm right or if I'm overlooking something, especially anyone who has the book? This shows up as an exercise in Nadler's Continuum Theory textbook where he does designate the $U$ to be open. However, I'm having trouble understand why such a thing would be a basis, especially when you consider $X_\alpha = X$ for every $\alpha \in D$ and all bonding maps to be identity functions.
The equation (I corrected an error in the equations) under 2 holds because whenever $\alpha_i \le \alpha$ we have $$f_{\alpha_i \alpha}\circ \hat{p}_\alpha = \hat{p}_{\alpha_i}$$ which follows immediately from the definitions. So they have the same inverse images of $U_{\alpha_i}$ and then we use the standard fact $(f\circ g)^{-1}=g^{-1} \circ f^{-1}$. The left hand side is then $\hat{p}^{-1}[\cap_{i=1}^n U_{\alpha_i}]$ so also in the base. This shows that the standard subbase for the inverse limit is in fact a base, getting the common upperbound $\alpha$ from the directedness. So the proof is correct ( I don't have the text but the idea of the proof is obvious).
Clearly all $U$ should be open in their respective spaces. Like Nadler also states.