# How do I implement an induction argument to show that we may assume that a partition $Q$ of $[a,b]\in{R}$ which refines $P$ has only one more point?

There is a lemma (32.2) in Kenneth A. Ross's Elementary Analysis text, that states:

Let $f$ be a bounded function on $[a,b]$. If $P$ and $Q$ are partitions of $[a,b]$ and $P\subseteq{Q}$, then $$L(f;P)\leq{L(f;Q)}\tag1,$$

where $L(f;P)$ and $L(f;Q)$ are lower sums for $f$ corresponding to $P$ and $Q$, respectively.

In his proof he states that an induction argument (which he assigns to an exercise) shows that we may assume that $Q$ has only one more point, namely $z$, than $P$.

What exactly would I supply for the induction argument?

In my own attempt at proving this lemma, I simply thought it'd be safe to assume that we may adjoin one point to $P$, to obtain a refinement $Q$. Proceeding with this assumption, it isn't hard to obtain $(1)$, and once that's been done, I can iterate the process, by refining Q with another partition that has only one point more than $Q$, and so on...

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There really is no (non-trivial) induction here. What you're trying to prove is "for all n, if Q is a refinement of P by n points, then (1) holds". Obviously (or as $\leq$ is transitive if you like) it suffices to prove this for n = 1 and then just repeatedly add points in. – Billy Dec 20 '12 at 5:04

If $Q$ has $n$ more points than $P$, then (1) holds