Context: In Chapter 13 of Spivak's Calculus, the Darboux Integral (although he does not refer to it as such) is introduced, which involves partitions:
Let $a \lt b$. A partition of the interval $[a,b]$ is a finite collection of points in $[a,b]$, one of which is $a$ and one of which is $b$.
In several of the examples that Spivak works out (e.g. the computation of $\displaystyle \int_0^b x dx$), a special type of partition, which I will denote as $\mathcal P_n$, is introduced. $\mathcal P_n$ is a partition that has $n$ equal subintervals. For example, suppose the interval of interest is $[0,4]$. Then $\mathcal P_4$ would denote the set $\{0,1,2,3,4\}$, $\mathcal P_8$ would denote $\{0,0.5,1.0,1.5,\cdots,3.5,4\}$.
For this post, the symbol $\mathcal P_n$ will be used in conjunction with the symbol $P'$, which simply denotes any arbitrary partition of a given $[a,b]$ (regardless of its partitioning structure).
I want to prove the following:
For an arbitrary $[a,b]$, where $a,b \in \mathbb R$, we can always construct a $P'$ partition such that for any $n$, $P' \not \subset \mathcal P_n$.
I think my approach is correct, but I suspect there is a much more direct approach. I proceed by using two proofs by contradiction.
Consider two cases.
Case 1: $a,b \in \mathbb Q$
Case 2: $a \in \mathbb R\setminus \mathbb Q$ or $b \in \mathbb R\setminus \mathbb Q$.
Case 1: $a, b \in \mathbb Q$
First, let $r_1 \lt r_2 \in P'$ and let $r_1,r_2 \neq a$ and $r_1,r_2 \neq b$. Additionally, let $r_1,r_2 \in \mathbb R\setminus \mathbb Q$. Suppose by contradiction that there exists an $n \in \mathbb N$ such that $P' \subset \mathcal P_n$.
Now, for any arbitrary $\mathcal P_n$, we have that any member of this partition can be expressed as $a+i\cdot\frac{b-a}{n}$ for any $i \in \{0,1,\cdots,n\}$. Then, in order for $P' \subset \mathcal P_n$, we must have that $r_1,r_2 \in \mathcal P_n$. Therefore, we know that:
\begin{align}r_1&=a+i_1\cdot \frac{b-a}{n} \\r_2&=a+i_2\cdot \frac{b-a}{n}\end{align}
Given that we stipulated $r_1,r_2 \in (a,b)$, we must have that $i_1,i_2 \in \{1,2, \cdots n-1\}$. It should be clear that the above two expressions force $r_1$ and $r_2$ to be members of $\mathbb Q$. Why? Because if $a,b \in \mathbb Q$, then $b-a=b+(-a) \in \mathbb Q$. Further, $\frac{i_1}{n}$ and $\frac{i_2}{n}$ are both also $\in \mathbb Q$. Therefore, $a+i_1\cdot \frac{b-a}{n} \in \mathbb Q$ and $a+i_2\cdot \frac{b-a}{n} \in \mathbb Q$. But we stated that $r_1, r_2 \notin \mathbb Q$. A contradiction.
So, if $a,b \in \mathbb Q$, then choosing any two irrational numbers between $a$ and $b$ as the $r_1$, $r_2$'s will produce the relevant $P'$.
Case 2: $a \in \mathbb R \setminus \mathbb Q$ or $b \in \mathbb R \setminus \mathbb Q$
Without loss of generality, suppose $a \in \mathbb R \setminus \mathbb Q$. Similarly to Case 1:
Let $r_1 \lt r_2 \in P'$ and let $r_1,r_2 \neq a$ and $r_1,r_2 \neq b$. This time, however, let $r_1,r_2 \in \mathbb Q$. Now, suppose by contradiction that there exists an $n \in \mathbb N$ such that $P' \subset \mathcal P_n$. Then we must have that $r_1,r_2 \in \mathcal P_n$. We can then argue as follows:
If $a \in \mathbb R \setminus \mathbb Q$ and $r_1 \in \mathbb Q$, then $r_1-a \in \mathbb R \setminus \mathbb Q$. Further, if $r_1,r_2 \in \mathcal P_n$, then $r_1-a$ must be a natural number multiple of $\frac{b-a}{n}$. Therefore, there is an $i \in \mathbb N$ such that: $r_1-a=i\cdot\frac{b-a}{n}$. Rearranging this as $\frac{i}{n}\cdot (b-a)$, and noting that $i \neq 0$, we must have that $b-a \in \mathbb R \setminus \mathbb Q$.
Next, consider the difference $r_2-r_1$. Because $r_1,r_2 \in \mathbb Q$, we must have that $r_2-r_1 \in \mathbb Q$. However, because $r_1,r_2\in \mathcal P_n$, we must also have that there exists an $i' \in \mathbb N$ such that: $r_2-r_1=\frac{i'}{n}(b-a)$. But because $b-a \in \mathbb R \setminus \mathbb Q$ and $\frac{i'}{n} \in \mathbb Q$, we must have that $r_2-r_1 \in \mathbb R \setminus \mathbb Q$. A contradiction.
A similar argument can be used for when $b \in \mathbb R \setminus \mathbb Q$.
In conclusion, there are partitions that include members that cannot be 'captured' by partitions of the form $\mathcal P_n$ (i.e. by $n$-equal-subinterval-partitions).