There seems to be an error in item (d) of Problem 1, Chapter 3 of Spivak's A Comprehensive Introduction to Differential Geometry, 3rd Edition, which I transcribed below.
There is something wrong with the definition of the functions $g_{i,j}$. Notice the $f_{i,j}$ functions are defined on $x_{i}(U_{i})\subset \mathbb{R}^{n}$, and $p\in M$ therefore the symbol $f_{i,j}(p)$ does not make any sense.
The immediate 'correction' would be to replace $f_{i,j}(p)$ with $f_{i,j}(x_{i}(p))$. But it would follow from this new definition that the pseudometric $\rho$ would actually be a metric on M, and then we could not hope to prove the properties we expect from $\rho$, considering the counterexample given in item (a).
Can somebody help me to find a proper correction to this error/typo?
(Problem 1): Let M be any set, and $\{(x_i,U_i)\}$ a sequence of one-one functions $x_i:U_i \to \mathbb{R}^N$ with $U_i \subset M$ and $x_i(U_i)$ open in $\mathbb{R}^n$, such that each $$x_j\circ x_i^{-1}:x_i(U_i\cap U_j) \to x_j(U_i \cap U_j)$$ is continuous. It would seem that M ought to have a metric which makes each $U_i$ open and each $x_i$ a homeomorphism. Actually, this is not quite true:
(a) Let $M = \mathbb{R} \cup \{\ast\}$, where $\ast \not\in \mathbb{R}$. Let $U_1=\mathbb{R}$ and $x_1:U_1 \to \mathbb{R}$ be the identity, and let $U_2=\mathbb{R}-\{0\} \cup \{\ast\}$, with $x_2:U_2 \to \mathbb{R}$ defined by $x_2(a)=a, \space a\not=0,\ast$ $x_2(\ast)=0$
Show that there is no metric on M of the required sort, by showing that every neighbourhood of $0$ would have to intersect every neighbourhood of $\ast$. Nevertheless, we can find on M a pseudometric $\rho$ (a function $\rho: M\times M \to \mathbb{R}$ with all properties for a metric except that $\rho(p,q)$ may be $0$ for $p\not=q$) such that $\rho$ is a metric on each $U_i$ and each $x_i$ is a homeomorphism:
(b) If $A \subset \mathbb{R}^n$ is open, then there is a sequence $A_1, A_2, A_3, \dots$ of open subsets of $A$ such that every open subset of A is a union of certain $A_i$'s.
(c) There is a sequence of continuous functions $f_i:A \to [0,1]$ with $support \space f_i \subset A$, which "separates points and closed sets": if C is closed and $p \in A-C$, then there is some $f_i$ with $f_i(p) \not\in f_i(A \cap C)$. Hint: First arrange in a sequence all the pairs $(A_i,A_j)$ of part (b) with $\overline{A_i} \subset A_j$.
(d) Let $f_{i,j}, j=1,2,3,\dots$ be such a sequence for each open set $x_i(U_i)$. Define $g_{i,j}:M\to [0,1]$ by
$g_{i,j}(p) = f_{i,j}(p), \space p \in U_i$ or $g_{i,j}(p) = 0, \space p \not\in U_i$
Arrange all $g_{i,j}$ in a single sequence $G_1,G_2,G_3, \dots$, let $d$ be a bounded metric on $\mathbb{R}$, and define $\rho$ on $M$ by:
$$\rho(p,q) = \sum_{i=1}^{\infty}\frac{1}{2^i}d(G_i(p),G_i(q))$$
Show that $\rho$ is the required pseudometric.
(e) Suppose that for every $p,q \in M$ there is a $U_i$ and $U_j$ with $p \in U_i$ and $q \in U_j$ and open sets $B_i \subset x_i(U_i)$ and $B_j \subset x_j(U_j)$ so that $p \in x_i^{-1}(B_i)$, $q \in x_j^{-1}(B_j)$, and $x_i^{-1}(B_i) \cap x_j^{-1}(B_j) = \emptyset $. Show that $\rho$ is actually a metric on M.
Thank you,
Gregory