|The Road to Reality
|Page 1 of 1|
|Author:||jbeckmann [ 26 Aug 2010, 16:43 ]|
|Post subject:||Exercise [16.13]|
Here is my proposal:
Exercises_16_13.pdf [23.92 KiB]
Downloaded 167 times
|Author:||deant [ 30 Apr 2011, 21:30 ]|
|Post subject:||Re: Exercise [16.13]|
JBeckmann, that's a very nice illustration how the first formulation of the "diagonal slash" argument (as given in the book) can be put into a similar-looking form as the second one. In particular, it shows exactly why it's called the "diagonal slash" argument (which isn't at all immediately evident in Penrose's first formulation of it).
However, as you remark at the end, there is nothing to prevent the argument being used for sets A with cardinality greater than aleph-0.
The most rigorous way to demonstrate that all such "diagonal slash" arguments are fundamentally the same (even when A has cardinality greater than aleph-0) is to formally define the "diagonal slash" procedure in the most general (or abstract) terms, and then show how the abstract formulation reduces to the particular ones when its parameters are chosen appropriately.
I've done this, and my solution is attached below.
The nature of the task required using mathematical language that's a lot more formal than that in the book, or even in most of the solutions posted on this forum... so I apologise to those readers who find it hard to understand. The examples on page 2 should help clarify the math-ese on page 1, or at least I hope they will!
|Page 1 of 1||Archived: 07 Aug 2014|
|Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group