[pwdisswordfish5]

> But we don't need to rely on Dijkstra's opinion at all.

I don’t. I refer to Dijkstra’s opinion because it happens to coincide with my opinion, it’s already written down, and it would be silly to spend effort explaining it over and over again.

[ModernMech]

But Dijkstra's opinion is not well founded, so it's definitely an appeal to authority. His opinion hinges on words like "ugly" and "preferred" without explaining what he means by this. He makes a choice of one over the other forms based on an unclear value function. His entire argument is this:

"There is a smallest natural number. Exclusion of the lower bound —as in b) and d)— forces for a subsequence starting at the smallest natural number the lower bound as mentioned into the realm of the unnatural numbers. That is ugly, so for the lower bound we prefer the ≤ as in a) and c). Consider now the subsequences starting at the smallest natural number: inclusion of the upper bound would then force the latter to be unnatural by the time the sequence has shrunk to the empty one. That is ugly, so for the upper bound we prefer < as in a) and d). We conclude that convention a) is to be preferred."

I mean... that doesn't exactly settle it. Sure, your opinion might coincide with that, but this is not a logical argument. First you're going to have to explain what unnatural numbers are (because they're not a thing as far as I can tell), and then you'll have to tell me what about them makes them "ugly", and how that makes the other option "preferable".

[alentist]

That you don't know what natural numbers are does not exactly inspire confidence in your commentary.

https://en.wikipedia.org/wiki/Natural_number

Dijkstra is saying the following.

Part 1:

Natural numbers have a least element X. (In fact, any subset of them does. This is called the well-ordering principle.)

Part 2:

Suppose lower bounds are exclusive. Then the lower bound L of any interval containing X must be less than X. Since X is the smallest natural number, L must be an unnatural (non-natural) number.

It is more convenient to stay in the naturals.

Therefore, it is more convenient to use inclusive lower bounds.

Part 3:

Suppose upper bounds are inclusive. Then, to represent an empty interval, the upper bound must be less than the lower bound. (If it were equal to the lower bound, we would get a singleton interval, not an empty interval).

It is already strange that we need an upper bound that is less than the lower bound. But there's more: When the lower bound is X, this would again take us out of naturals.

Therefore, it is more convenient to use exclusive upper bounds.

Conclusion:

It is more convenient to use inclusive lower bounds and exclusive upper bounds, i.e. convention (a).

This argument is basically airtight.

[ModernMech]

I know what natural numbers are. But "unnatural" numbers is not a term people use.

"It is more convenient to stay in the naturals."

Right here, you do the same thing Dijkstra does. You declare that one way is more "convenient" than the other (he calls it ugly), without defining what "convenient" means. This makes the argument subjective. What you think is convenient, others may not think so. Or, one may agree with you on a matter of convenience, but would prefer a different choice due to some other priority e.g. learnability.

So I just don't agree this is an airtight argument. It's a subjective argument for a preference at best.

[alentist]

> You declare that one way is more "convenient" than the other (he calls it ugly), without defining what "convenient" means. This makes the argument subjective.

Nope.

1. Some things are objectively more convenient than others.

2. I've explained exactly what I mean. And so did Dijkstra.

Since you apparently don't want to use natural numbers for indices, what do you want to use?