I'm reading Spivak Calculus, and in Part III, Chapter 13 Integrals, I have come across to lower bounds, lower sums, upper bounds and upper sums. In page 216 Spivak attempts to prove the lemma
If $Q$ contains $P$ (i.e., if all points of $P$ are also in $Q$), then $$L (f,P) \leq L(f,Q) \\U(f,P) \geq U(f,Q)$$
He considers a special case for proving, "consider a special case in which $Q$ contains just one more point than $P$". Then he makes a figure like this:
and then he writes
The Set $\{ f(x) : t_{k-1} \leq x \leq t_k\}$contains all the numbers in $\{ f(x) : t_{k-1} \leq x \leq u\}$ and possibly some smaller ones, so that the greatest lower bound of the first set is less than or equal to the greatest lower bound of the second set.
I don’t know how just more number of points ensure that the greatest lower bound will be less. Is it true (I mean what he wrote) for every case or just for the case he was explaining? For the case which he is explaining it is obvious but will it be true in every case?
Best Answer
it's always true. Adding points that are larger than the greatest lower bound can't make the lower bound higher because it must be still be a lower bound for the original points that were in the set.
And adding points that are lower then the greatest lower bound will force the greatest lower bound to be smaller because the greatest lower bound can only be at most as big as these new lower points.
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Suppose $X$ is a set. And suppose the greatest lower bound of $X$ is $r$.
Now suppose $k \not \in X$. What is the greatest lower bound of $X \cup \{k\}$?
Well case 1: If $k \ge r$ then for every $r$ is less than or equal to every $x\in X$. And $r\le k$ so $r$ is less than or equal to every $x \in X\cup \{k\}$ so $r$ is still a lower bound. And its the greatest lower bound of $X$ so any $w> r$ isn't a lower bound of $X$ so there will be an $x_1\in x$ so that $x_1< w$. But $x_1 \in X\cup\{k\}$ so $w$ isn't a lower bound of $X\cup \{k\}$ either.
So the greatest lower bound is still $r$.
Case 2: If $k < r$ then $r$ is no longer a lower bound of $X\cup \{k\}$ because $k \in X\cup \{k\}$ is $k < r$. So any lower bound must be smaller.
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