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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?

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    why the downvote? It seems to me a legitimate question. For some reason questions asking for general recipes always get downvotes – 2017-01-21
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    I didn't downvote, but please note that there is nothing to *show* in cases 1. and 2. If you know that a set is an interval, you can *determine* its borders, but no need to show that the right border is the supremum. – 2017-01-21
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    of course we know that.. that is not the point. They are the classic exercises that are given in any Calculus and Analysis courses. Of course we know it, but if we had to start university asking people to focus only on the Riemann Zeta function, we wouldn't go much further – 2017-01-21
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    I don't understand the question about 2. -- $b$ is the supremum, since by definition, for all $x \in A$, $x \le b$, and if $x \le c$ for all $x \in A$, then $b \le c$ because the set of all such $c$ is $[b,\infty)$. Thus I think the most interesting case is 3. But it's so general, I don't think there can be a general recipe, that is part of the whole challenge of mathematics, the set of all proofs is not computable (i.e. it is not possible to prove everything using an algorithm i.e. recipe). – 2017-01-21
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    For instance, the question of whether or not the supremum of the set of possible values of some quantity is finite is equivalent to whether or not that quantity is bounded, and my experience/memory seems to suggest that such proofs are often rather ad hoc, i.e. have to be found with luck, ingenuity, skill, experience, talent, etc. It's not possible to prove all theorems using an(y single) algorithm. – 2017-01-21
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    @William yeah it is a basic proof. But it doesn't come straight from the definition. For example $1> 0$ , but that doesn't come from the definition of $1$ or $0$, you need to prove it, although that's a very basic proof – 2017-01-22
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    Not necessarily an algorithm. An algorithm should being you to the solution, a recipe guides you through a solution. For example the one in case $1$ is an algorithm, but the one in case two is probably a recipe! As well as the one in case 3, indeed it doesn't tell you the reasoning you have to do, but shows you where you need to look at, what you should be looking for and aiming to obtain. What is the general strategy. – 2017-01-22
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    @Euler_Salter I think it's an interesting question. Still, I don't think there's even a general strategy for case 3 -- you have to try to exploit the properties of the given set in an ad hoc manner. I could be wrong, but that is my impression. – 2017-01-22
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    @William tell me what you think about this, it looks like a working recipe for case $2$ – 2017-01-22

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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 $b = \sup(A)$

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    Yes this looks correct and completely rigorous -- it is definitely better than the argument I gave. – 2017-01-22
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    @WIlliam Thank you, I wonder now if it is possible to find the equivalent for infimum, but I reckon it is just the same, using the AP in the opposite way – 2017-01-22
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    I would use the fact that $\sup A = -\inf \{-x: x \in A \}$ -- then I think you can transfer all of the results fairly immediately. See: https://en.wikipedia.org/wiki/Infimum_and_supremum#Duality – 2017-01-22
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    true, I forgot about that! Now the only one left is case three, which should reduce into one of these cases anyway, hopefully – 2017-01-22