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

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?

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  • $\begingroup$ 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? $\endgroup$ – Matthew Towers Sep 16 at 21:13
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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.

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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.

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