Why in this definition of Riemann integral isn't required that the mesh gets smaller?

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Why in this definition for the Riemann integral, taken from wikipedia page for it, isn't explicitly required that the mesh gets smaller? Because a refinement of a partition doesn't require that the mesh gets smaller. That means that I can keep one interval the same width and refine the other part.

For all $ε > 0$, there exists a tagged partition $y_0, ..., y_m$ and $r_0, ..., r_{m − 1}$ such that for any tagged partition $x_0, ..., x_n$ and $t_0, ..., t_{n − 1}$ which is a refinement of $y_0, ..., y_m$ and $r_0, ..., r_{m − 1}$, we have

$${\displaystyle \left|\sum_{i=0}^{n-1}f(t_{i})(x_{i+1}-x_{i})-s\right|<\varepsilon .}$$

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The point is that it has to hold for any refinement, including those where you do make the mesh smaller and smaller.