Tradeoff between different formulations of partition of unity

In Wikipedia, a partition of unity of a topological space $X$ is a set $R$ of continuous functions from $X$ to the unit interval $[0,1]$ such that for every point, $x\in X$,

• there is a neighbourhood of $x$ where all but a finite number of the functions of $R$ are $0$, and
• $\;\sum_{\rho\in R} \rho(x) = 1$.

The existence of partitions of unity assumes two distinct forms:

• Given any open cover $\{U_i\}_{i\in I}$ of a space, there exists a partition $\{\rho\}_{i\in I}$ indexed over the same set $I$ such that $\textrm{supp } \rho_i\subset U_i$. Such a partition is said to be subordinate to the open cover $\{U_i\}_i$.

• Given any open cover $\{U_i\}_{i\in I}$ of a space, there exists a partition $\{\rho_j\}_{j\in J}$ indexed over a possibly distinct index set $J$ such that each $\rho_j$ has compact support and for each $j\in J$, $\textrm{supp } \rho_j\subset U_i$ for some $i\in I$.

According to Wikipedia,

thus one chooses either to have the supports indexed by the open cover, or compact supports. If the space is compact, then there exist partitions satisfying both requirements.

Would anybody give a simple example (probably in $\mathbb{R}$) showing the difference between the two different choices mentioned above?

Take a cover of $\mathbb{R}$ by two open sets, $U_0 = (-\infty,2)$ and $U_1 = (-2,\infty)$ . Then let $$\rho_0(x)= \begin{cases} 1 & x\leq -1 \\ \frac{1}{2}-\frac{1}{2}x & -1 \leq x \leq1 \\ 0 & x\geq 1 \end{cases}$$ $$\rho_1(x)= \begin{cases} 0 & x\leq -1 \\ \frac{1}{2} + \frac{1}{2}x & -1 \leq x \leq 1 \\ 1 & x\geq 1 \\ \end{cases}$$
Then $\{\rho_i\}_{(0,1)}$ is subordinate to $\{U_i\}_{(0,1)}$ in the first sense but not the second.
Note that the cover can't possibly have a partition of unity subordinate to it with compact supports, since $U_0 \setminus U_1$ is not compact, but $\rho_0$ must be $1$ on that set.
Here is a trivial example. Let $U=\mathbb{R}$. Then $\{U_0\}$ with $U_0=U$ is an open cover of $U$. $\{\rho_0\equiv1\}$ gives a partition with the same index. But there is no way to find $\{\rho_0\}$ with $\textrm{supp }\rho_0\Subset U_0$ since $\rho_0=1$ on $U$.