If for an interval $[a,b]$ you have two partitions $Z_1: a = x_0 < \cdots < x_m = b$ and $Z_2: a = y_0 < \cdots < y_n = b$ where the mesh of $Z_1$ is strictly greater than the one of $Z_2$, i.e. $$\max \{y_i - y_{i-1}\mid i=1,\ldots,n\} < \delta < \max \{x_i - x_{i-1}\mid i=1,\ldots,m\}$$ then for every $i \in\{1,\ldots,n\}$ there is a $j \in \{0,\ldots,m\}$ such that $$[y_{i-1},y_i] \subset (x_{j-1},x_j)$$ or $$[y_{i-1},y_i] \subset (x_{j-1},x_j] \cup [x_j, x_{j+1})$$ with $(x_{-1},x_0]=:\emptyset:=[x_m,x_{m+1})$
I don't understand why this holds, with no limit as to how small the "spacing" between individual $x_i$ can be, wouldn't you be able to have a subinterval $[y_{i-1},y_i]$ with $y_{i-1} - y_i < \delta$ of $Z_2$ which in $Z_1$ corresponds to three (or really arbitrarily many) small subintervals? What am I missing here? Thank you for your help.
Edit: For context I encountered this in the lecture notes of a first semester real analysis course in a proof of: Let $f: [a,b]: \to \mathbb{R}$ be a regulated function (earlier defined as a function for which for all $\epsilon > 0$ there exists a step function $φ_\epsilon$ such that $|f(x)-φ_\epsilon(x)| \leq \epsilon$ for all $x\in [a,b]$). Then for every $\epsilon > 0$ there exists a $\delta > 0$ such that for partitions $Z = \{x_0,...,x_n\} $ $$mesh(Z) < \delta \implies |\int_a^b f(x)dx - S(f,Z)| < \epsilon$$ ($S(f,Z) $ is defined as the Riemann sum over f with respect to $Z$.)