How does more number of points ensure a lower "greatest lower bound"?

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 $P$.


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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Tags:

Integration

Calculus

Real Analysis

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