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(Completely unrelated to the previous blogpost.)

This post is an experiment, in a way. When teaching mathematics or when writing certain kinds of professional communications (journal articles, but also course notes or reference works) one often seeks to reduce duplication by stating and proving general results. Often these take a form that is much more abstract than the intended applications, and one positive side is that by removing specific features that are irrelevant to the chain of logical reasoning in the proof, one avoids the danger of “not seeing the wood for the trees”.

However, there is the risk that by distilling what one perceives as the key argument into a minimalist and abstract form, one loses both the original context and the motivation for the particular hypotheses chosen.

Of course, there is no right or wrong side here; one needs to allow both perspectives. But I wondered whether the following example, which arises from distilling an argument I cooked up in some recent attempts at research, looks too abstract and artificial, or whether readers might find the hypotheses and objects relate to “natural” examples they encounter in their own research.

In what follows $\mathbb N$ denotes the set of natural numbers, starting from $1$ (apologies to any passing set theorists). As a parallel experiment, I’ve tried to be a little more detailed than I would be as a “working analyst”, so that what follows could be read by students who still wish to see i’s dotted and t’s crossed.

Let $(\mathcal S,\preceq)$ be a partially ordered set (which, for the purposes of what follows, one should think of as uncountable). Suppose $\mathcal S$ has the following properties:

(F) if $A, B\in \mathcal S$ then there exists $D\in \mathcal S$ with $A\preceq D$ and $B\preceq D$;

(σ) if $A_1\preceq A_2 \preceq \dots$ is an increasing sequence in $\mathcal S$, then there exists $C\in \mathcal S$ with $A_n\preceq C$ for all $n\in\mathbb N$.

(Here I am using the analyst’s convention that “increasing” does not mean “strictly increasing”; I am not sure if this is standard usage in the setting of posets.)

Now suppose we have a function $h: {\mathcal S} \to {\mathbb R}$ which is monotone in the sense that $A\preceq B$ implies $h(A)\leq h(B)$, and bounded above in the sense that $\sup_{A\in\mathcal S} h(A) < \infty$.

Claim: $h$ attains its supremum at some element of $\mathcal S$. That is, there exists $C\in {\mathcal S}$ such that $h(C) = \sup_{A\in \mathcal S} h(A)$.

Proof. Let $K=\sup_{A\in\mathcal S} h(A)$. Pick $A_1\in \mathcal S$. We inductively construct $A_2, A_3, \dots$ in $\mathcal S$ which satisfy $A_n \preceq A_{n+1}$ and $h(A_{n+1}) > K- 1/n$ for all $n\in \mathbb N$.

For the inductive step: given $n\in \mathbb N$ and $A_n\in\mathcal S$, pick some $B_n\in \mathcal S$ such that $h(B_n) > K - 1/n$ (possible by the definition of $K$) and then use the hypothesis (F) to obtain $A_{n+1}\in \mathcal S$ satisfying $A_n \preceq A_{n+1}$ and $B_n \preceq A_{n+1}$; monotonicity of $h$ ensures that $h(A_{n+1})\geq h(B_n) > 1- 1/n$.

Having obtained this sequence $(A_n)_{n\in\mathbb N}$, the hypothesis (σ) ensures that there exists $C\in \mathcal S$ such that $A_n\preceq C$ for all $n\in \mathbb N$. Then for each $n\in\mathbb N$ we have $K \geq h(C) \geq h(A_n) > K-1/n$, and we conclude that $h(C)=K$ as required. QED.

Question for readers: if you saw this result in a paper or a book that you were reading, followed by a single application which is much more concrete and specialized, would you feel happy with this? Or would you prefer to see the original argument as it occurred “in the wild”, followed by a remark that there is a more “Platonic” ideal form that could be formulated?

Bonus points, by the way, if you manage to guess the original setting for the argument which gave rise to what’s written above.

3 Comments leave one →
1. 7 December, 2020 6:09 pm

My vote (usually): Prove it in the wild! We all know how to generalize.

If I want to talk about some dimension function on separable subspaces of some huge space (or whatever that is), it is better I simply talk about that, no need to introduce a new spooky h on some abstract poset. In a case where I can’t resist, I might add a remark: “note that all used was….”. Even when you *do* have another application coming up it is sometimes clearer to prove the first instance in the wild, and in the second application just note that the same proof applies. Finally, I am not sure that “cleaning the clutter” always makes for a cleaner exposition. To the contrary, sometimes the general idea is already clear in a certain instance, and everyone can see the generalization without needing to write it down.

Sometimes it turns out to be worthwhile to formulate more generally, but it’s hard to know that in real time.

2. Matthew Daws permalink
8 December, 2020 9:45 am

I think this is in the eye of the beholder, but in this particular instance, I am strongly inclined to agree with Orr. Maybe I’ve been doing Analysis too long, but this sort of argument doesn’t seem “tricky” enough for me to need to separate out the details. That said, I do agree with the general principle though.
Bonus question: I think this is from some construction like the full group Cstar algebra. You are taking the supremum over a set of representations, and you want to argue that as the set is closed under infinite direct sums, supremums are attained.

3. 11 December, 2020 5:29 pm

Thanks Orr and Matt. I must admit that the process of writing the blog post more or less swayed me already towards the option both of you chose. To be fair, I don’t think I would actually have included the argument stated above in a research paper; it was more of a deliberately extreme case.

I guess I was wondering if some of the Online Categorical People would turn up to say that this is some general principle about posets, analogous to the pigeonhole principle.

Matt is on the right lines concerning the “bonus question”, although my intended application is more pedestrian. I may get round to posting an update as a separate post.

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