# Using partitions of unity to construct a nonzero continuous function on $(0,1)$

Say I have an open cover of the interval $$(0,1)$$. I'll call this open cover $$\{V_j\}_{j \in J}$$. On each of these open sets there is a continuous, nonzero, complex-valued function $$f_j: V_j \to \mathbb{C} \setminus \{0\}$$. Is it possible to find a partition of unity $$\{p_j\}_{j \in J}$$ subordinate to the open cover $$\{V_j\}_{j \in J}$$ such that the continuous complex-valued function $$\sum_j p_j\, f_j$$ is never zero?

If it's possible but not easy to see, then I wouldn't mind if someone just provides reference(s)?

## 1 Answer

It's not always possible. Here's a counterexample. Take $$V_1 = (0,\frac23)$$, $$V_2 = (\frac13,1)$$, $$f_1 \equiv 1$$, and $$f_2\equiv -1$$. Then, whatever $$p_1$$ and $$p_2$$ are, we have $$\sum_j p_j f_j = p_1 - p_2$$, which is real-valued. Since $$p_1-p_2 \equiv p_1\equiv 1$$ on $$(0,\frac13)$$ and $$p_1 - p_2 \equiv -p_2 = -1$$ on $$(\frac23,1)$$, by the intermediate value theorem there must be a point $$x\in (\frac13,\frac23)$$ where $$p_1(x)-p_2(x)=0$$.