Sometimes when I'm doing proofs regrading Riemann Integrals I would like to create an explicit parition whose norm is less than $ \delta $ . I'm familiar with creating the following partition of $ [a,b] $ : $ \prod = \{ a < a+\frac{a}{b-a}<a+\frac{2a}{b-a} <...<a+\frac{i\cdot a}{b-a} < ....< a+\frac{n \cdot a}{b-a} \} $ here the mesh of partition is $ \max\{ | x_i - x_{i-1} | : i = 1,...,n \} = \frac{a}{b-a} $
But I would like an example for a partition $ \prod = \{ x_0 < x_1 < ... < x_n \} $ of $ [a,b] $ whose mesh satisfies $ \max\{ | x_i - x_{i-1} | : i = 1,...,n \} < \delta $ ( where $ \delta > 0 $ ). ( In the above example, if $ a \neq 0 $ then it doesn't necessarily mean that $ \frac{a}{b-a} < \delta $, but I can't think of an example for a partition that can satisfy of mesh being less than $ \delta $ )
Thanks in advance for any help!
Edit ( + solution as given in the comments ):
I made a mistake in the partition above, Here's an example for a correct partition of $[a,b] $:
$ \prod = \{ a < a+\frac{1\cdot (b-a)}{n}<a+\frac{2\cdot (b-a)}{n} <...<a+\frac{i\cdot (b-a)}{n} < ....< a+\frac{n\cdot (b-a)}{n} = b \} $
Now, as stated in the comments, looking at the mesh of this partition, it is $ \frac{b-a}{n} $ and I want $ \frac{b-a}{n} < \delta $ so I see that $ \frac{b-a}{\delta} < n $. So we'll choose $ N \in \mathbb{N} $ s.t. $ \frac{b-a}{N} < \delta $ ( the argument for this is by Archimedean Principle ), thus, if I want to create a partition of mesh less than $ \delta $ I can create the following partition of $ [a,b] $:
$ \tilde{\prod} = \{ a < a+\frac{1\cdot (b-a)}{N}<a+\frac{2\cdot (b-a)}{N} <...<a+\frac{i\cdot (b-a)}{N} < ....< a+\frac{N\cdot (b-a)}{N} = b \} $ and the mesh of this partition is $ \frac{b-a}{N} < \delta $, as wanted.