A Question about a Proof

The Student

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Apr 25, 2012
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My hope is to understand something about proofs even though I understand the concept that is being proved. My notes have,

Lemma 5.1 (Partition Refinement): If P and Q are partitions of [a, b] such that
Q ⊃ P, then
L(P, f) ≤ L(Q, f) ≤ U(Q, f) ≤ U(P, f):

It is sufficient to prove the lemma when Q contains just one more point
than P. (Why?)


And then the notes go on to prove the statement.

So, I am having trouble understanding why one extra point is sufficient for Q ⊃ P. I can see how one extra point supports the claim by being consistent with the original statement, but what about the rest of the possible subsets of Q?
 
I'm not sure I would call it "mathematical induction" but it certainly is similar. An important point here is that a "partition" of a set consists of a finite number of points. Suppose Q has "n" more points than P. Let Q1 be P with a single point of Q added, Q2 be Q1 with one more point of Q added, etc. up to Qn= Q. If we have proved the lemma for "one more point", then it is true for Q1 so it is also true for Q2, etc.
 
I'm not sure I would call it "mathematical induction" but it certainly is similar. An important point here is that a "partition" of a set consists of a finite number of points. Suppose Q has "n" more points than P. Let Q1 be P with a single point of Q added, Q2 be Q1 with one more point of Q added, etc. up to Qn= Q. If we have proved the lemma for "one more point", then it is true for Q1 so it is also true for Q2, etc.

Oh, thank-you, that makes sense.
 
Just one small disagreement with HallsofIvy and that may be because I don't understand the under lying premises. The disagreement is with the statement "a 'partition' of a set consists of a finite number of points". As I was taught ( a very long time ago), a partition of a set is just a (generally countable) division of that set so that the union of any two sets of the partition is the empty set and the union of all the sets of the partition is the original set. Thus the disagreement is with the finite aspect of the statement. If we assume that the original set is countable [as the statement of the problem implies], all we need to do is replace the finite of HallsofIvy's with countable and the same argument goes through.
 
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