tag:blogger.com,1999:blog-4987609114415205593.post8928334517698948593..comments2020-07-13T20:26:53.037+01:00Comments on M-Phi: Average Height of a BeatleJeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-4987609114415205593.post-88019172865946468142013-07-20T22:49:48.658+01:002013-07-20T22:49:48.658+01:00Thanks Jeff. I'm curious to see how this turns...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. Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-17732819730285253782013-07-20T21:43:16.854+01:002013-07-20T21:43:16.854+01:00I'm happy to talk tenselessly, a la Quine, but...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".<br />Now 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.<br /><br />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 ...<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-60956008299706683302013-07-20T21:35:08.690+01:002013-07-20T21:35:08.690+01:00[i]But maybe B(20 July 2013) = {Paul, Ringo}.[/i]
...[i]But maybe B(20 July 2013) = {Paul, Ringo}.[/i]<br /><br />John is no longer a Beatle but Ringo is? Why is that? You seem to be digging yourself in deeper here.<br /><br />What is the average height of the members of the Roman senate? They're all dead but the question is still sensible. <br /><br />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.<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-59145418757562586542013-07-20T13:58:31.879+01:002013-07-20T13:58:31.879+01:00:)
'So, when I came to Oxford in the early 196...:)<br />'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).Jeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-16282496341740944832013-07-20T08:25:58.200+01:002013-07-20T08:25:58.200+01:00Jeff, you're not living up to your usual stand...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 ;-)Robert Blacknoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-41214602203802142332013-07-20T05:05:29.226+01:002013-07-20T05:05:29.226+01:00Hey, I'm building up to something! (To do with...Hey, I'm building up to something! (To do with applicability)<br />There are no Beatles, but I was doing it tenselessly ... <br />Here we go with a time-indexed version<br />B(t) = {x | x is a Beatle at time t}<br />and then let t = 20 July 1967, say.<br /><br />But maybe B(20 July 2013) = {Paul, Ringo}.<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-79106260205645411752013-07-20T04:16:24.355+01:002013-07-20T04:16:24.355+01:00I'm a little confused. Is there a point to thi...I'm a little confused. Is there a point to this? Was the second half of the post omitted?<br /><br />In any event, the Beatles are defunct and there are no Beatles. Therefore the expression 1/|B| is undefined.Anonymousnoreply@blogger.com