0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Oct 24 '18 at 14:04
  • $\begingroup$ @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. $\endgroup$ – Ramandeep Singh Arora Oct 24 '18 at 16:32
0
$\begingroup$

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 $f_i(x)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.