Cauchy criterion for Riemann integrability 
I have a serious concern regarding the proof of the Cauchy criterion implying the Riemann integrability $\text{Cauchy Criterion}:$ $\space \space$ A function: $[a,b] \to \mathbb{R}$ belongs to $\mathcal{R}[a,b]$ if and only if for every $\epsilon>0$ there exists $n_\epsilon>0$ such that if $\dot P$ and $\dot Q$ are any tagged partitions of $[a,b]$ with $||\dot P||<n_\epsilon$ and $||\dot Q||<n_\epsilon,$ then $$|S(f;\dot P)-S(f;\dot Q)|<\epsilon$$
  The forward implication is clear.Conversely
  
  $(\Leftarrow)$ For each $n\in \mathbb{N},$ let $\delta_n>0$ be such that if $\dot P$ and $\dot Q$ are tagged partitions with norms $<\delta_n,$ then $$|S(f;\dot P)-S(f;\dot Q)|<\frac{1}{n}.$$
  Evidently we may assume that $\delta_n\ge \delta_{n+1}$ for $n\in \mathbb{N};$ otherwise, we replace $\delta_n$ by $\delta'_n:=\min\{\delta_1,..,\delta_n\}.$
  $\space$ For each $n\in \mathbb{N},$ let $\dot P_n$ be a tagged partition with $||\dot P_n||<\delta_n$. Clearly, if $m>n$ then both $\dot P_m$ and $\dot P_n$ have norms $<\delta_n$, so that $$|S(f;\dot P_n)-S(f;\dot P_m)|<\frac{1}{n} \space \space \space ,m>n \tag1$$
  And the proof goes ..

Question
How can we assume that $\delta_n\ge \delta_{n+1}?$ I get the whole point of assuming it because only then can we get $(1)$ as a Cauchy sequence. But how can we assume it?.. What would go wrong if $\delta_n<\delta_{n+1}?$ Also this whole proof is implying that $\epsilon \to 0$ implies $\delta \to 0$. Thinking about this graphically makes sense, but nowhere in the definition is this obvious to me.
 A: I'll answer your question about how we can assume $\delta_{n} \geq \delta_{n+1}$. (You say that you understand why we want it to be true.)
Let's fix $n$ for the moment. Let's say we already found $\delta_1,\ldots, \delta_n$. According to the hypothesis, we can find $\delta_{n+1}^{\ast} > 0$ such that 
$$
|S(f,P) - S(f,Q)| < \frac{1}{n+1}
$$
for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$.
I used the notation $\delta_{n+1}^{\ast}$ because it is our first try for $\delta_{n+1}$. It could be that $\delta_{n+1}^{\ast}$ is larger than $\delta_{n}$. 
To fix that, for our second attempt we define
$$
\delta_{n+1} = \min(\delta_n,\delta_{n+1}^{\ast}).
$$ 
Let's check that $\delta_{n+1}$ does what we want.
Clearly $\delta_{n} \geq \delta_{n+1}$. So that's taken care of. 
We also want $\delta_{n+1}$ to have the property that 
$$
|S(f,P) - S(f,Q)| < \frac{1}{n+1}
$$
for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$. Let's check it.
From the definition of $\delta_{n+1}$, it's clear that $\delta_{n+1}^{\ast} \geq \delta_{n+1}$. So for any tagged partitions $P$ and $Q$ with $\|P\| < \delta_{n+1}$ and $\|Q\| < \delta_{n+1}$, we have $\|P\| < \delta_{n+1}^{\ast}$ and $\|Q\| < \delta_{n+1}^{\ast}$, and therefore (by the original property of $\delta_{n+1}^{\ast}$) we have
$$
|S(f,P) - S(f,Q)| < \frac{1}{n+1}.
$$
