$B := \{x \mid x$ is a Beatle $\}$and let:

$h_m(x) = r$stand for the quantity function which maps human beings to their heights in

*metres*.

Then define:

$\overline{h}_B := \frac{1}{|B|} \sum_{x \in B} h_m(x)$This value is

*the average height of a Beatle*.

Let

*C*be a class of human beings. Then:

$\overline{h}_C := \frac{1}{|C|} \sum_{x \in C} h_m(x)$This value is

*the average height of an element of C*.

I'm a little confused. Is there a point to this? Was the second half of the post omitted?

ReplyDeleteIn any event, the Beatles are defunct and there are no Beatles. Therefore the expression 1/|B| is undefined.

Hey, I'm building up to something! (To do with applicability)

ReplyDeleteThere are no Beatles, but I was doing it tenselessly ...

Here we go with a time-indexed version

B(t) = {x | x is a Beatle at time t}

and then let t = 20 July 1967, say.

But maybe B(20 July 2013) = {Paul, Ringo}.

Cheers,

Jeff

Jeff, you're not living up to your usual standards of pedantry. You can't divide a real number by a cardinal number: you've got to go via the canonical embedding of the finite cardinals into the reals ;-)

ReplyDelete:)

ReplyDelete'So, when I came to Oxford in the early 1960s there was a lot of pedantry around ... Or, rather, I should say there was a lot of pedanticness.' (Jerry Cohen).

[i]But maybe B(20 July 2013) = {Paul, Ringo}.[/i]

ReplyDeleteJohn is no longer a Beatle but Ringo is? Why is that? You seem to be digging yourself in deeper here.

What is the average height of the members of the Roman senate? They're all dead but the question is still sensible.

In any event I'm still not clear as to why you posted this. Surely most readers are familiar with the concept of the arithmetic mean.

I'm happy to talk tenselessly, a la Quine, but it just complicates matters, somewhat needlessly! So, we'd just relativize the predicates to times, and then work with those. E.g., "x is a member of the Roman Senate at time t".

ReplyDeleteNow I think about it, you're right: Ringo and Paul are (i.e., now) former Beatles; so, I guess that $B(t) = \varnothing$, for $t$ after 1969-ish.

The unstated reason for posting it has got to do with category theory, structural set theory and applicability; but I actually need to sort out in my mind how such a theory deals with non-abstract objects. And I want a few examples to refer back to at that future point ...

Jeff

Thanks Jeff. I'm curious to see how this turns out. It's a long way from the arithmetic mean of a finite set to category theory.

ReplyDelete