Why the Largest Width Matters
A partition breaks a closed interval into smaller subintervals. Those widths need not all be equal: a nonuniform partition may have several narrow pieces and one much wider piece. To describe how finely the whole interval has been divided, it is not enough to look at the average width. We need to keep track of the largest width.
This largest width is called the mesh of the partition. It gives a simple measure of the partition's coarsest part: every subinterval is no wider than the mesh, and at least one subinterval has exactly that width. When a partition is refined, new division points are added. Refinement cannot increase the mesh, although it does not always make the mesh strictly smaller.
Definition of Mesh
Let \(a<b\). A partition \(P\) of \([a,b]\) is a finite ordered list of points
The subintervals determined by \(P\) are \([x_{i-1},x_i]\), for \(i=1,\ldots,n\). The width of the \(i\)th subinterval is \(\Delta x_i=x_i-x_{i-1}\), which is positive because the partition points are strictly increasing.
The maximum exists because a partition has finitely many subintervals. Thus \(\|P\|\) is one of the actual widths, not an average or an estimate. Every width satisfies \(0<\Delta x_i\leq\|P\|\), and at least one width equals \(\|P\|\).
Computing Mesh for a Nonuniform Partition
Worked Example: Finding the Mesh from Unequal Widths
Consider the partition \(P=\{0,\frac14,1,\frac54,2\}\) of \([0,2]\). First calculate each consecutive difference:
The largest of these four widths is \(\frac34\). Therefore,
Check: Each listed width is at most \(\frac34\), and the second and fourth widths equal \(\frac34\). The calculated mesh is therefore the largest subinterval width, as required.
This example also shows why the average width is not the mesh. The total interval length is \(2\), divided into four pieces, so the average width is \(\frac12\). But two pieces have width \(\frac34\), making the mesh \(\frac34\), not \(\frac12\).
Uniform Partitions as a Special Case
A partition is uniform when every subinterval has the same width. For a uniform partition of \([a,b]\) into \(n\) subintervals, each width is \(\frac{b-a}{n}\). Since all widths are equal, this common width is also the mesh.
Worked Example: Mesh of a Uniform Partition
Divide \([-1,2]\) into \(6\) equal subintervals. The total length is \(2-(-1)=3\), so the common width is
To display the partition points, start at \(-1\) and repeatedly add \(\frac12\):
There are six consecutive widths, each equal to \(\frac12\). Hence
Check: The first and last partition points are \(-1\) and \(2\), and the six widths add to \(6\cdot\frac12=3\), the length of the interval. Since every width is \(\frac12\), the largest width is \(\frac12\).
What It Means to Refine a Partition
A partition \(Q\) of \([a,b]\) is a refinement of a partition \(P\) if every point of \(P\) is also a point of \(Q\). In other words, refinement adds partition points without removing the original ones. It subdivides some of the original subintervals, while leaving the others unchanged.
A refinement is called proper if it adds at least one new point. Adding points can make the partition more detailed, but a proper refinement does not necessarily reduce the mesh. The reason is that refinement may add points only inside intervals that are already narrower than the widest interval.
Worked Example: A Proper Refinement with Unchanged Mesh
Let \(P=\{0,1,2\}\). Its two widths are
so \(\|P\|=1\). Now take \(Q=\{0,\frac12,1,2\}\). It contains all the points of \(P\), together with the new point \(\frac12\), so \(Q\) is a proper refinement of \(P\). Its widths are
Thus
Check: The refinement added a point, and the two smaller widths are each at most \(1\), but the interval \([1,2]\) was left intact with width \(1\). It still attains the maximum, so the mesh did not decrease.
Refinement Cannot Increase the Mesh
Although refinement may leave the mesh unchanged, it cannot make the mesh larger. To see why, consider any one subinterval of the refined partition. Since all the old partition points are retained, that new subinterval lies inside one of the old subintervals. Its width is therefore no greater than the width of the old subinterval containing it.
Let \([y_{j-1},y_j]\) be any subinterval of \(Q\). Because \(Q\) contains every point of \(P\), there is no point of \(P\) strictly between \(y_{j-1}\) and \(y_j\). Consequently, this refined subinterval lies within some subinterval \([x_{i-1},x_i]\) of \(P\).
Containment gives \(x_{i-1}\leq y_{j-1}<y_j\leq x_i\). Subtracting the left endpoint from the right endpoint shows that \(y_j-y_{j-1}\leq x_i-x_{i-1}\).
By definition, \(x_i-x_{i-1}\leq\|P\|\). Thus every width \(y_j-y_{j-1}\) in \(Q\) is at most \(\|P\|\).
Since every subinterval width in \(Q\) is at most \(\|P\|\), their maximum is at most \(\|P\|\). Therefore \(\|Q\|\leq\|P\|\), which proves the theorem.
The proof establishes a non-strict inequality. For a strict decrease, every subinterval of \(P\) whose width equals \(\|P\|\) must be subdivided so that none of its resulting pieces still has that full width. If even one widest subinterval is left unchanged, the refined partition still has mesh \(\|P\|\).
Worked Example: Refining the Widest Subintervals
Use \(P=\{0,\frac14,1,\frac54,2\}\) from the first example, whose mesh is \(\frac34\). Refine both intervals of width \(\frac34\) by inserting their midpoints. The widest intervals are \([\frac14,1]\) and \([\frac54,2]\), with midpoints \(\frac58\) and \(\frac{13}{8}\). The resulting partition is
The consecutive widths are
The largest of these is \(\frac38\), so \(\|Q\|=\frac38\). Therefore, the mesh has decreased from \(\frac34\) to \(\frac38\).
Check: The original points \(0,\frac14,1,\frac54,2\) all remain in \(Q\), confirming that \(Q\) is a refinement. Each old widest interval was split into two pieces of width \(\frac38\), and the other widths are \(\frac14\). Hence no width exceeds \(\frac38\), and some widths equal it.
Making Mesh as Small as Needed
The mesh gives a precise way to express that every subinterval is small. For example, the statement \(\|P\|<\varepsilon\), where \(\varepsilon>0\), means that every subinterval width is less than \(\varepsilon\). For a uniform partition of \([a,b]\) into \(n\) pieces, this condition becomes \(\frac{b-a}{n}<\varepsilon\). Choosing \(n\) sufficiently large makes the uniform mesh as small as desired.
Worked Example: Choosing a Uniform Partition with a Required Mesh
Choose a uniform partition of \([0,3]\) whose mesh is strictly less than \(\frac14\). With \(n\) equal subintervals, the mesh is \(\frac{3}{n}\). We require
Since \(n\) is positive, multiplying by \(4n\) gives \(12<n\). The smallest integer satisfying this strict inequality is \(n=13\). For that choice,
because \(12<13\). A choice of \(n=12\) would give \(\frac{3}{12}=\frac14\), which does not satisfy the strict requirement. Thus thirteen equal pieces suffice, and twelve do not.
Check: The interval length is \(3\), so thirteen equal widths sum to \(13\cdot\frac{3}{13}=3\). Each width is \(\frac{3}{13}\), which is strictly less than \(\frac14\), as required.
A Common Pitfall: Confusing Refinement with Smaller Mesh
A frequent mistake is to conclude that a proper refinement must have strictly smaller mesh. The earlier example shows why that conclusion fails: new points can be added while at least one widest interval remains unsplit. The correct general conclusion is \(\|Q\|\leq\|P\|\), not necessarily \(\|Q\|<\|P\|\).
For instance, starting from \(P=\{0,1,2\}\), it would be wrong to argue that inserting \(\frac12\) automatically makes the mesh smaller. The actual calculation is
The correction is to inspect every resulting width, especially the intervals that previously attained the maximum. To ensure a strict decrease, subdivide all widest intervals into narrower pieces.
Why Mesh Is Useful
Partition mesh summarizes the worst-case width with one number. In a nonuniform partition, a small average width cannot rule out a wide subinterval; the mesh can. When partitioning an interval to obtain finer spatial or time resolution, requiring a small mesh ensures that no part of the interval has been left with a large gap. This is why mesh, rather than average width alone, is useful when comparing how finely partitions divide an interval.
Keep three facts distinct: the mesh is the maximum width; a refinement retains every old partition point and cannot increase the mesh; and strict decrease requires that the old widest intervals no longer remain intact at their full width.
Check Your Understanding
Use the definitions and results above to answer the following questions.
- For \(P=\{0,\frac12,\frac32,2\}\), calculate each subinterval width and find \(\|P\|\).
- Let \(P=\{0,2,4\}\) and \(Q=\{0,1,2,4\}\). Is \(Q\) a refinement of \(P\), and does its mesh strictly decrease?
- Explain why every subinterval of a refinement \(Q\) lies inside an old subinterval of \(P\).
- A uniform partition of \([0,5]\) has \(20\) subintervals. What is its mesh?
- How can a proper refinement leave the mesh unchanged?
