So, I was reading A.S Schwarz paper, The genus of a fiber space. There was a statement "Any locally finite open covering $\{U_i\}_{i \in I}$ of a normal space $X$ has a system of continuous real-valued functions $\{f_i\}_{i \in I}$ from $X$ such that
(a) $0 \leq f_i \leq 1$,
(b) $f_i(x)=0$ if $x \not\in U_i$,
(c) for each $x \in X$ there exist $i \in I$ such that $f_i(x)=1$."
I know that "Any locally finite open covering $\{U_i\}_{i \in I}$ of a normal space $X$ has a partition of unity, say $\{\phi_i\}_{i \in I}$, subordinate to that cover". The partition of unity $\{\phi_i\}_{i \in I}$ satisfy the properties (a) and (b) already and also $\Sigma_{i \in I} \phi_i(x)=1$ (or $\Sigma_{i \in I} \phi_i(x)>0$ both are equivalent). To have property (c) I need to modify my partition of unity (that's what I think, there could be another solution to this problem).
Since my covering is locally finite, each $x \in X$ belongs to finitely many elements of the covering, say $U_1, \ldots, U_n$. Then, define $U_x=U_1 \cup \ldots \cup U_n$ and $\phi_x=\phi_1+\ldots +\phi_n$. Then, I will get a open covering $\{U_x\}_{x \in X}$ and a system of functions $\{\phi_x\}_{x \in X}$ satisfying (b) and (c). But, now the problem is I have changed the indexing set.
I need my system of functions with the same indexing set. I would be really helpful if you could guide me in this problem.
Best Answer
It turns out the $\{f_i\}_{i \in I}$ are the functions which after normalization leads to the partition of unity.
Suppose $\{U_i\}_{i \in I}$ is a locally finite open covering of a normal space $X$ then there exist an open covering $\{V_i\}_{i \in I}$ of $X$ such that $\bar{V_i} \subset U_i$. Observe that the covering $\{V_i\}_{i \in I}$ is also locally finite, thus there exist an open covering $\{W_i\}_{i \in I}$ of $X$ such that $\bar{W_i} \subset V_i$. Since $\bar{W_i}$ and $X\setminus V_i$ are closed disjoint subsets of $X$, by Urysohn's lemma, there exist a function \begin{equation*} f_i:X \rightarrow [0,1] \end{equation*} such that $f_i(X \setminus V_i)=0$ and $f_i(\bar{W_i})=1$. Also $f_i^{-1}(\mathbb{R} \setminus 0) \subset V_i$, we have \begin{equation*} \bar{W_i} \subset \text{support}(f_i) \subset \bar{V_i} \subset U_i. \end{equation*} Moreover, for each $x \in X$ there exist $i \in I$ such that $x \in W_i$, since $\{W_i\}_{i \in I}$ is a covering, and hence $f_i(x)=1$.
Thus, any locally finite open covering $\{U_i\}_{i \in I}$ of a normal space $X$ has a system of continuous real-valued functions $\{f_i\}_{i \in I}$ from $X$ satisfying the conditions: (a) $0 \leq f_i \leq 1$; (b) support$(f_i) \subset U_i$; (c) at each point $x \in X$ there exist $i \in I$ such that $f_i(x)=1$.