This seems like a silly question, but I think it will help me to understand the partition of unity.
Suppose that $(U_\alpha)_{\alpha \in I}$ be arbitrary open cover of manifold $M$ (or topological space) with partition of unity subordinate $(\psi_\alpha)_{\alpha \in I}$. In most cases I've seen, $\forall \alpha$, $\exists x \in U_\alpha$ such that $\psi_\alpha (x)=1$. I want to know if there are any examples such that $\exists \alpha$ (or $\forall \alpha$), $\forall x \in U_\alpha$ such that $\psi_\alpha(x)<1$.
This can certainly happen. First consider the open cover of $\Bbb R$ given by $\{ I_{2k}\colon k\in \Bbb Z\}$, where $I_n = (k-1.1,k+1.1)$. Let $\{\psi_{2k}\colon k\in \Bbb Z\}$ be a corresponding partition of unity. (Note that $\psi_{2k}(x)=1$ for all $x\in[k-0.9,k+0.9]$, since those points are contained in $I_{2k}$ but in no other $I_n$.)
But now change the open cover to $\{ I_n\colon n\in \Bbb Z\}$, so that there's much more overlapping. Then $\{ \frac12\psi_{2k}(x)\colon k\in\Bbb Z \} \cup \{ \frac12\psi_{2k}(x-1) \colon k\in\Bbb Z \}$ is a partition of unity every function in which is bounded above by $\frac12$, and $\frac12\psi_{2k}(x)$ is supported in $I_{2k}$ while $\frac12\psi_{2k}(x-1)$ is supported in $I_{2k+1}$.