Time Complexity 3

Currently, there are four things I want to make clear in this topic.
1) The Piazza question regarding the proof of Lemma 3.6 still gets me confused, and I want to clarify this during OH.

2) Again lemma 3.6 (and this means proofs to show T(n) is non-decreasing). The proof denotes n by k instead. The third case tries to show that for k > 2, P(1), P(2), ..., P(k-1) implies P(k). I tried the proof itself several times and I still don't agree that "Since P(k-1) holds, to prove that P(k) holds as well, it suffices to prove that T(k-1) <= T(k)." I suppose that (if an error were to exist, it would have to be here)   the proof goes on to show that since P(k-1) holds: "for every positive integer m, if m < k - 1, then T(m) <= T(k-1)", then T(m) <= T(k-1) <= T(k), the last inequality needed because we are trying to show P(k), which says: "for every positive integer m, if m < k, then T(m) <= T(k)", not merely T(k - 1). Of course this would seem silly because T(k-1) <= T(k) relies on that which we are trying to prove, that: T(k) is non-decreasing.

3) Danny skimmed over an intuition by using summation(a/(b^d)) to understand the master theorem. I read the hefty (I imagine, part-proof) of the theorem in the course notes (At least I think it was here), but I couldn't see a clear connection.

4) In class, we moved on to apply the Master Theorem to more interesting examples, in particular for multiplying n-bit numbers Gauss-wise (of which I now understand after some review) and closest point pairs (of which I don't understand completely. The slides are not posted, yet - but once they are I plan to go over them with scrutiny.)

Meanwhile, I'm hoping that more examples come up on the Master Theorem, because I'm not entirely confident on the choice of a: "The number of pieces for subproblems/the number of recursive calls." It seems that only experience can aid me in understanding this.

EDIT: Most excellent! In regards to the above:
1) I see that by the definition of non-decreasing, m < n implies T(m) < T(n), but this is equivalent to m <= n implies T(m) <= T(n). This is because if m = n, then obviously T(m) = T(n).
2) If you look at the final sentence above, I think that we are making a logical loop by needing T(k-1) <= T(k), but now I understand that this is exactly what the Vassos was trying to prove. Proving T(k-1) <= T(k) is two-fold, because for one, we break out of the logical loop and two, we are able to cover the "final case" which I was so caught up on (Since we are trying to show P(k - 1), but needing to prove T(k - 1) <= T(k) as well, since P(k - 1) cover all cases (induction hypothesis cases) before that).