The context of this question is in relation to the paragraph of text below from David Bachman's "A Geometric Approach to Differential Forms":
The part that seems contradictory to me is that as $\max\{\Delta x_i\}\rightarrow0$, $n$ would necessarily need to approach infinity. This is because, intuitively, for the size of the largest interval length to approach $0$, there must be an infinite number of $x_i$ (in the limit,... bit handwavey).
This is what I have done so far; finding a contradiction by assuming that $n$ remain fixed (to try and suggest that $n$ cannot be fixed).
Define (this is the questionable assumption): \begin{equation}\tag{1} I_\Delta=\{1,2,\dots,n-1\},\;n\geq 2 \end{equation} Define: \begin{equation}\tag{2} x_1, x_2, \dots, x_n\in [a,b],\;x_i< x_{i+1},\;\forall i\in I_\Delta \end{equation} More specifically, choose $x_1=a,x_n=b$.
Define $\Delta=\{\Delta x_i\}=\{\Delta x_i=x_{i+1} - x_{i}\;|\;i\in I_\Delta\}$
Since all these definitions are used to set up our intuition of splitting an interval $[a,b]$ however we please, we must require that: \begin{equation}\tag{3} \sum_{i\in I_\Delta}\Delta x_i=b-a \end{equation} Let $h(C)=C=\max\Delta$. This implies that: \begin{equation}\tag{4} \Delta x_i\leq C,\;\forall i\in I_\Delta \end{equation} From (2), $x_i<x_{i+1}\implies \Delta x_i =x_{i+1}-x_i>0\implies 0\leq \Delta x_i$. Combining (4) with (2), we get that: \begin{equation}\tag{5} 0\leq \Delta x_i\leq C,\;\forall i\in I_\Delta \end{equation} For all $\Delta x_i\in \Delta$:
Let $\lim\limits_{C\rightarrow 0}h(C)=\lim\limits_{C\rightarrow 0}C=0=L$:
Using the squeeze theorem on (5), where $g(C)=0,\;f(C)=\Delta x_i,\;h(C)=C$: \begin{equation*} \lim\limits_{C\rightarrow 0}f(C)=L=0\implies \lim\limits_{C\rightarrow 0}\Delta x_i=0 \end{equation*} \begin{equation}\tag{6} \therefore \lim\limits_{\max\Delta\rightarrow 0}\Delta x_i=0,\;\forall \Delta x_i\in \Delta \end{equation} Taking the limit of (3) we get: \begin{align}\tag {3 lim} \lim\limits_{\max \Delta \rightarrow 0}\left(\sum_{i\in I_\Delta}\Delta x_i\right)=\lim\limits_{\max \Delta \rightarrow 0}\left(b-a\right)=b-a \end{align} But we also know that: \begin{align*} \lim\limits_{\max \Delta \rightarrow 0}\sum_{i\in I_\Delta}\Delta x_i&= \sum_{i\in I_\Delta}\lim\limits_{\max \Delta \rightarrow 0}\Delta x_i=0\neq b-a \end{align*} which contradicts (3 lim) as $b-a\neq 0$. I suspect that the only way for this to be consistent with requirement (3) is if as $\max\Delta\rightarrow 0,\;n\rightarrow \infty$.
Is there a flaw in this logic?
In the summation above how many $\Delta x_i$ we have? is it changing? or fixed? it could not be fixed. The idea could be clearer if you think about this in the regular approach in analysis i.e. for any $\epsilon >0$ there exists a partition of $[a,b]$, $x_1,\cdots x_n$ such that $max \Delta x_i < \epsilon$. In analysis if we want to prove something approach zero we use for any delta small we can make the value of what we have smaller. Hope this helps