I am trying to understand and visualize this following theorem to get a better understanding about what it means and how its proof works. Here is the theorem:

Theorem: Let $A\subset\mathbb{F}$ and $\alpha\in\mathbb{F}$. $\alpha=\text{lub}A$ iff $(\alpha,\infty)\cap{A}=\emptyset$ and for all $\epsilon>0, (\alpha-\epsilon,\alpha]\cap{A}\neq\emptyset.$ Similarily, $\alpha=\text{glb}A$ iff $(-\infty,\alpha)\cap{A}=\emptyset$ and for all $\epsilon>0, [\alpha,\alpha+\epsilon)\cap{A}\neq\emptyset.$

If possible, I would like to see a full intuistic explanation to better my understanding and visualization of this theorem.


The first statement can be rephrased as "$\alpha$ is an upper bound for $A$", that is, for any $x \in A$, $x \leq \alpha$. The second statement can be rephrased as "there can be no smaller upper bounds for $A$ than $\alpha$". This is because if you take $\beta=\alpha-\varepsilon$ for any $\varepsilon>0$, there is a member of $A$ larger than $\beta$, so $\beta$ is not an upper bound. Everything is reversed for the infimum.

Visually, there are two cases. The supremum can be just the largest member of $A$. For instance, $\sup [0,1]=1$. If there is no maximum, then the supremum is the only upper bound for which there is a sequence in $A$ which converges to it. For instance, $\sup (0,1)=1$ because any member of $(0,1)$ is at most $1$ and because the sequence $\{ 1-1/n \}_{n=2}^\infty$ converges to $1$ from inside $(0,1)$. Again everything is reversed for the infimum.

A different visual answer is that the supremum is the maximum member of the closure, where we take the closure in the extended real numbers (so the closure of a set which is not bounded above contains $+\infty$). This maximum will always actually exist.


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