# Vector Calculus: Mesh Size and Size of Largest Rectangle

1. Jan 26, 2015

1. The problem statement, all variables and given/known data
I need to find a sequence of partitions , let's call it S of R=[0,1]x[0,1] such that as the number of partitions k→∞ , then limit of the area of the largest subinterval of the rectangle in the partition, denoted a(S) tends to 0, but the mesh size m(S) is a non-zero value.

2. Relevant equations

3. The attempt at a solution

If I denote Δx_i as the partition width for the i-th interval and say Δx_i = (i2/k2)-((i-1)2/k2) = (2i-1)/(k2). Then the width of the largest subinterval will approach 0 as k→∞, which in turn means the the area of the largest subinterval will go to 0. However, I'm unsure how to show the mesh size for the partition can't be 0? Any help is appreciated.

2. Jan 26, 2015

### LCKurtz

3. Jan 26, 2015

You mentioned thinking about long skinny rectangles, and the only thing that came to mind is the dirac-delta function which has a size equal to unity. I don't know if that helps my case, but nothing else spurs into my noggin at the moment.

4. Jan 26, 2015

### LCKurtz

How small can the mesh size of a partition having a long single long skinny rectangle be? Say the rectangle has height $1$ and (narrow) width $w$.

5. Jan 26, 2015

Would the mesh size just be $w$ ?

6. Jan 26, 2015

### LCKurtz

What is the definition of mesh size? Suppose your mesh was just this one rectangle? What would its mesh size be by that definition?

7. Jan 26, 2015

The mesh size is the greatest widthed partition in the interval [a,b]?

8. Jan 26, 2015

### LCKurtz

Remember this is a 2D partition. You are dividing a 2d region into rectangles, not dividing an interval [a,b] into subintervals. These rectangles form a partition of an area. The term "width" isn't really appropriate. Think in terms of largest "diameter" of the rectangle. So what would be the mesh size of the single rectangle I mentioned?

9. Jan 26, 2015

Apparently you are just guessing. That sentence doesn't make any sense to me. I didn't ask about the area of that rectangle, and even if I did, the area of a $1$ by $w$ rectangle certainly isn't $w^2$. Please reread my quoted post and answer the last question.