# Variant of Partition of Unity

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.

• There must be another property your $\phi_i$ satisfy. Otherwise they could all be identically $0$ and satisfy (a) and (b). And your attempt would be doomed. – Henno Brandsma Oct 24 '18 at 14:04
• @HennoBrandsma Yeah you are right. I forgot to mention earlier the partition of unity also satisfy $\Sigma_{i \in I} \phi(x)=1$, which now I have mentioned after editing my question. Thanks. – Ramandeep Singh Arora Oct 24 '18 at 16:32

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 $$f_i(x)=1$$.