General Method to prove $s$ is a supremum of the given interval

Question

Is it possible to work out a "general" recipe to show an element, say $s$, is the supremum (or infimum) of a given set?

I've noticed that in my lecture notes and in the book I'm reading, I can never find a general recipe, but surely there must be guidelines. There are three cases (I think):

1. Show $b$ is the supremum of $A$ when $A =[a,b]$ or $A=(a,b]$
2. Show $b$ is the supremum of $A$ when $A=[a,b)$ or $A = (a,b)$
3. Show $b$ is the supremum of $A$ when $A$ is a set described by a given property, for example $A = \{n\in \mathbb{N}\,\,\,\,\, | \,\,\,\,\,\,(-1)^nn^2\}$ (I've made this up).

For example I show there is a general recipe for case 1:

Case 1 Since we are looking for the supremum, it doesn't really matter whether it is $A =[a,b]$ or $A=(a,b]$ , so I'll just use $A=(a,b]$. By definition of the set, we know that $b \geq x$, $\forall x\in A$. Now suppose there exists another element $c$ such that $c\geq x$, $\forall x \in A$, then since $b\in A$ we have that $b \leq c$ and hence $b = \sup(A)$.

Case 3 Well here the general recipe is very general, meaning that it depends a lot on the description of the set. However one can generalise saying that first of all, if possible, we should try to rewrite the expression giving us the elements of the set. After this, we should start from remembering that $n\in\mathbb{N}$ means that $n \geq 1$ or $n>0$. Similarly, we can use domains and properties of several functions, such as $-1\leq \cos(x) \leq 1$. We use these inequalities to try to bound the expression describing the elements of $A$. Once we have found that, we are basically either in case $1$ or in case $2$. Sometimes, we might have to use induction, but that is rare. Should we use approximation property for suprema?

Case 2 Is there a general recipe? I always make these exercises quite...luckily, although I can feel there is some kind of underlying structure. Most of the times I used Archimedean properties, if not all the times.

MY QUESTION

How would we define the general recipe for case $2$? Can someone tell me? And also, are my recipes for case $1$ and case $3$ correct?

Answer 1

I've finally found the answer myself, I knew there was one for case $2$. Here it is:

Suppose we have $A = [a,b)$ and we want to show $b = \sup(A)$.

1. We have $\forall x\in A$, $a \leq x < b$ hence by definition of $A$, we have that $b$ is an upper bound for $A$.
2. Now suppose $u$ is another upper bound for $A$. I want to show by contradiction that $b \leq u$. SO suppose $b > u$. Then take $\epsilon := b-u >0$. By the corollary of Archimedean Property, we know that $\exists n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} < \epsilon$. But this gives $$\frac{1}{n_0} < b-u$$ which again gives $$ u < b - \frac{1}{n_0}$$ Now notice that since $n_0\in \mathbb{N}$, we have $n_0 >0$ and so $\frac{1}{n_0} >0$, which tells us that $b - \frac{1}{n_0} \in A$ Hence $u$ is not an upper bound for $A$. So by contradiction $b \leq u$

So by $1$ and $2$ we know that $b = \sup(A)$