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Partition of an Interval

Partition of an Interval

A division of an interval into a finite number of sub-intervals. Specifically, the partition itself is the set of endpoints of each of the sub-intervals.

 

Example: {0, 0.2, 0.9, 1.1, 1.6, 2} partitions [0,2] into sub-intervals [0,0.2], [0.2,0.9], [0.9,1.1], [1.1,1.6], [1.6,2].

 

 

See also

Norm of a partition

Key Formula

P={x0,x1,x2,,xn}wherea=x0<x1<x2<<xn=bP = \{x_0, x_1, x_2, \ldots, x_n\} \quad \text{where} \quad a = x_0 < x_1 < x_2 < \cdots < x_n = b
Where:
  • PP = The partition, written as a set of points
  • [a,b][a, b] = The original interval being partitioned
  • x0,x1,,xnx_0, x_1, \ldots, x_n = The partition points, listed in increasing order
  • nn = The number of sub-intervals created by the partition

Worked Example

Problem: Create a partition of the interval [0, 6] into 4 sub-intervals, and identify each sub-interval. Then find the mesh (norm) of the partition.
Step 1: Choose 4 sub-intervals by selecting 3 interior points between 0 and 6. The points do not need to be equally spaced. Let's pick 1, 3, and 4.
P={0,1,3,4,6}P = \{0, 1, 3, 4, 6\}
Step 2: List the 4 sub-intervals formed by consecutive partition points.
[0,1],  [1,3],  [3,4],  [4,6][0, 1], \; [1, 3], \; [3, 4], \; [4, 6]
Step 3: Compute the width of each sub-interval.
Δx1=10=1,Δx2=31=2,Δx3=43=1,Δx4=64=2\Delta x_1 = 1 - 0 = 1, \quad \Delta x_2 = 3 - 1 = 2, \quad \Delta x_3 = 4 - 3 = 1, \quad \Delta x_4 = 6 - 4 = 2
Step 4: The mesh (or norm) of the partition is the width of the largest sub-interval.
P=max(1,2,1,2)=2\|P\| = \max(1, 2, 1, 2) = 2
Answer: The partition P = {0, 1, 3, 4, 6} divides [0, 6] into 4 sub-intervals: [0, 1], [1, 3], [3, 4], and [4, 6]. The mesh of this partition is 2.

Another Example

Problem: Create a regular (uniform) partition of [2, 5] into 6 equal sub-intervals.
Step 1: Find the common width of each sub-interval by dividing the total length by the number of sub-intervals.
Δx=526=36=0.5\Delta x = \frac{5 - 2}{6} = \frac{3}{6} = 0.5
Step 2: Generate each partition point by starting at the left endpoint and repeatedly adding Δx.
x0=2,  x1=2.5,  x2=3,  x3=3.5,  x4=4,  x5=4.5,  x6=5x_0 = 2, \; x_1 = 2.5, \; x_2 = 3, \; x_3 = 3.5, \; x_4 = 4, \; x_5 = 4.5, \; x_6 = 5
Step 3: Write the partition as a set.
P={2,  2.5,  3,  3.5,  4,  4.5,  5}P = \{2, \; 2.5, \; 3, \; 3.5, \; 4, \; 4.5, \; 5\}
Answer: The regular partition is P = {2, 2.5, 3, 3.5, 4, 4.5, 5}, creating 6 sub-intervals each of width 0.5.

Frequently Asked Questions

Does a partition have to use equally spaced points?
No. A partition only requires that the points be in strictly increasing order from a to b. When all sub-intervals have equal width, it is called a regular (or uniform) partition, but unequal widths are perfectly valid.
How is a partition related to Riemann sums?
A Riemann sum approximates the area under a curve by summing the areas of rectangles built on each sub-interval of a partition. As the mesh of the partition approaches zero (sub-intervals get narrower), the Riemann sum converges to the definite integral. The partition is the foundational structure that makes this approximation possible.

Regular (Uniform) Partition vs. General Partition

A regular partition spaces all points equally, so every sub-interval has the same width Δx = (b − a)/n. A general partition allows sub-intervals of different widths. Regular partitions are simpler to compute with, but the formal definition of the Riemann integral uses general partitions to ensure convergence regardless of how the interval is divided.

Why It Matters

Partitions are the building blocks of integral calculus. Every Riemann sum, trapezoidal rule, and Simpson's rule calculation begins by partitioning an interval into sub-intervals. Understanding partitions lets you see how the definite integral is constructed as a limit of increasingly fine subdivisions.

Common Mistakes

Mistake: Forgetting that the partition must include both endpoints of the original interval.
Correction: The first point must equal a and the last point must equal b. Without both endpoints, the sub-intervals would not cover the entire interval [a, b].
Mistake: Confusing the number of partition points with the number of sub-intervals.
Correction: If the partition has n + 1 points (including a and b), it creates n sub-intervals. For example, 5 points produce 4 sub-intervals.

Related Terms

  • IntervalThe domain being divided by the partition
  • MeshLargest sub-interval width in a partition
  • Riemann SumUses partitions to approximate integrals
  • Definite IntegralDefined as limit of sums over partitions
  • Trapezoidal RuleNumerical method built on partitions
  • FiniteA partition always has finitely many points