tag:blogger.com,1999:blog-3891434218564545511.post6483341480835292095..comments2024-03-28T19:56:42.305-05:00Comments on Alexander Pruss's Blog: More amusement with infinitesimal probabilitiesAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-3891434218564545511.post-5405635319973339142012-09-20T15:28:33.894-05:002012-09-20T15:28:33.894-05:00DL:
That sounds right, and that's why the pro...DL:<br /><br />That sounds right, and that's why the problem doesn't happen when we work with half-open intervals.<br /><br />Still, the argument does show that an intuitive scaling property does not hold once we bring in infinitesimals.<br /><br />The argument is making me have vague thoughts about how we shouldn't think of a continuum as made up of points. Years ago, when I was a math grad student, my dad pushed against the idea that a continuum is made of points. As a mathematician, I couldn't think of a continuum in any other way. I still can't really, but these kinds of cases make me think that Dad might well have been right that a different way to think is needed, and are giving me glimpses.<br /><br />Maybe I should look at John Bell's work on continua again.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-47051994919534065922012-09-20T12:49:25.077-05:002012-09-20T12:49:25.077-05:00I think the problem is we're not being careful...I think the problem is we're not being careful with our infinitesimals. Consider the discrete set [0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1] being broken into [0 to 0.5] and [0.5 to 1]. Each "half" contains 6 elements, but the whole is 11 elements, not 12, because of the double-counting of 0.5. That double-counting is handled in subtracting the P({½}) but it's not accounted for in saying that P([0, ½]) is half of P([0, 1]): it's really half-plus-an-infinitesimal-amount. That extra amount is the missing discrepancy.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-61941846372981461422012-09-20T11:38:10.083-05:002012-09-20T11:38:10.083-05:00Interesting. You might have a system like that. ...Interesting. You might have a system like that. I need to think about it. But normally infinitesimals are members of a field that extends the reals. And for the members of a field, if a=a+b, then b=0. So what I said is true for hyperreals, formal Laurent series, and other constructions of infinitesimals.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-44624393168411297352012-09-20T11:33:39.192-05:002012-09-20T11:33:39.192-05:00You write, "The only way this can be is if P(...You write, "The only way this can be is if P({1/2})=0, contrary to (1)."<br /><br />But is that right? I don't know exactly how infinitesimals are defined, but take an analogy with transfinite arithmetic. Aleph_0 = Aleph_0 + 1. One might think that the only way this equation can work is if 1 = 0, but one would be wrong. Why not think that infinitesimals work in a similar way? Perhaps the relation of an infinitesimal quantity to an ordinary finite quantity is like the relation of an ordinary finite quantity to an infinite one.Jonathan Livengoodhttps://www.blogger.com/profile/08264815112941067048noreply@blogger.com