# A condition for greatest lower bound - Intuition

In page 2 of this lecture note, it states that one of the conditions for $$w = \inf A$$ is that $$\forall r \in \mathbb{R}, r>w \rightarrow \exists a \in A, a.

I totally understand its contrapositive: if $$s$$ is a lower bound for A, then $$w\geq s$$.

However, I don't get the intuition behind the condition I stated first. What does having an element in set $$A$$ that is smaller than a number bigger than $$\inf A$$ have to do with the "greatest" lower bound?

• the condition says that anything bigger than w isn't a lower bound for A. Isn't that an "intuitive" property for a glb to have? – Matthew Towers Sep 16 at 21:13

In general nested quantifiers like this can be thought of as challenge response games. For instance, this one says: $$w$$ is the infimum IFF for all legal moves $$r$$ the adversary makes, where legal means $$r > w$$, I have a winning move, namely some $$a \in S$$ with $$a < r$$.
Then to build intuition, it may help to draw some pictures of sets (intervals, blobs of points on a line), and play this game a few times. E.g. draw the infimum $$w$$ so that you can quickly see which moves $$r$$ are legal, and then convince yourself that you have a winning response for each one.
Recall that if $$A$$ is a nonempty subset of $$\mathbb{R}$$ and it is bounded below, then its infimum, say, $$w \equiv \inf A$$ exists in $$\mathbb{R}$$. The infimum of $$A$$ must satisfy two conditions:
1. It is no larger than any given element of $$A$$, i.e., it is a lower bound of $$A$$: $$\forall x(x \in A \implies w \leq x)$$
2. It is the greatest lower bound of $$A$$, meaning no other real number $$r$$ exceeding $$w$$ can be a lower bound, so if $$r$$ is a lower bound of $$A$$, then $$r \leq w$$.
So given $$r > w$$, if we couldn't find some $$x \in A$$ with the property that $$w \leq x < r$$, then either $$x < w$$ for all $$x \in A$$, contrary to our choice of $$w$$ (indeed, contrary to assumption 1); or $$w < r \leq x$$, for all $$x \in A$$, meaning $$r$$ is a lower bound of $$A$$ which is larger than $$w$$, contrary to our second assumption. We are thus forced to conclude that if $$r > w$$, we can always find some $$x \in A$$ such that $$w \leq x < r$$, as desired.