How does more number of points ensure a lower "greatest lower bound"? 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?
 A: What he wrote is correct.  If $Q$ contains $P$ the greatest lower bound of $Q$ is less than or equal to the greatest lower bound of $P$.  It is not about the number of points, it is about containing.  If $Q$ contains $P$ two things can happen, analogous to adding one more point.  Either no point of $Q \setminus P$ is lower than the greatest lower bound of $P$, or at least one point of $Q \setminus P$ is lower than the greatest lower bound of $P$.  In the first case the greatest lower bound of $Q$ is the same as the greatest lower bound of $P$.  In the second case it is lower because it can be no greater than the point that is lower than all the points of $P$.  Either way, the greatest lower bound of $Q$ is less than or equal to the greatest lower bound of $P$.
A: 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.
=====
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. 
...
A: The claim is always true. It is a relatively simple concept, but is important enough to warrant a careful explanation. Hopefully by the end of this answer you'll agree the claim is obvious ^_^
What is the minimum of $A = \{5,4,3\}$? Compare this to the minimum of $B = \{5,4,3,2\}$. It is clear that the minimum of $B$ is $2$, while the minimum of $A$ is $3$. So $B$ has a smaller minimum.
Let's move one step more abstract. Say $A = \{5,4,3\}$ and $B = \{5,4,3,x\}$. What can we say about $\min A$ vs $\min B$? Well, $\min A = 3$ again, and we have two cases for $x$. If $x < 3$, then we see $\min B = x < 3$. Otherwise, $\min B = 3$. Notice the addition of $x$ can only make the minimum smaller.
Let's move yet another step further. If $A$ is any old set of reals, and $B = A \cup \{x\}$, then we can run the argument from before in the abstract. We know that $\min A$ is some number, and $x$ can only make $\min B$ smaller, since every element of $A$ is also in $B$. Thus $\min B \leq \min A$
Finally, whenever $A \subseteq B$, we have $\min B \leq \min A$, since $B$ has more elements than $A$, and thus has more chances to be small. In case all the stuff in $B \setminus A$ is bigger than $\min A$, the worst we can do is have $\min B = \min A$, since $\min A \in A$ is also in $B$!
The exact same logic applies to $\inf$ instead of $\min$ once you pass to limits.

I hope this helps ^_^
