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I am trying to read Smullyan A Beginner's Guide to Mathematical Logic and that Chapter $2$, Infinite Sets has left me quite bewildered!

Problem 1. (page $18$)

If a set A has exactly three elements, how many subsets of A are there (including A itself and the empty set)? What about a 5-element set? In general, if a set A has n elements, then in terms of n, how many subsets of A are there?

Solution (Page 24) is not talking anymore about set A, but about $I_{n}$ with subsets of S;

Particularly confusing is

" Thus if $S_{1}$ ,......, $S_{k}$ are the subsets of $I_{n}$, then the subsets of $I_{n+1}$ are the $2k$ sets... "

$2k$ sets? What could all this mean? If anyone has that book in hand to look at that Solution 1 and to explain it thoroughly for me, I would be very grateful!

I might just skip that chapter of course but I suspect it is not a very good idea ...

Aili J.
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  • See number-of-subsets-of-a-set-having-r-elements. – Mauro ALLEGRANZA Dec 11 '16 at 14:48
  • $I_n$ is the set ${ 1,2, \ldots, n }$; it has $n$ elements. – Mauro ALLEGRANZA Dec 11 '16 at 14:51
  • He shows that the number of subsets if $I_{n+1}$ is twice that the number of subsets of $I_n$. Thus, if $I_n$ has $k$ subsets, then $I_{n+1}$ has $2k$ subsets. – Mauro ALLEGRANZA Dec 11 '16 at 14:53
  • Then he starts from $I_1 = { 1 }$ : $1$ element and $2$ subsets : $\emptyset$ and $I_1$ itself. It follows that $I_2$ will have twice as much subsets, i.e. $2 \times 2=4=2^2$ subsets. – Mauro ALLEGRANZA Dec 11 '16 at 14:54
  • Please, note that up to now we are dealing with finite sets. – Mauro ALLEGRANZA Dec 11 '16 at 15:00
  • @MauroALLEGRANZA Does that original set A (in the question) have then 6 subsets? And next 5-element set - 25 subsets? – Aili J. Dec 11 '16 at 15:06
  • If $A$ has five ($5$) elements it has $2^5=32$ subsets : $I_1$ has $2$, $I_2$ has $2 \times I_1=4$, $I_3$ has $2 \times I_3=8$, $I_4$ has $2 \times I_3=16$... – Mauro ALLEGRANZA Dec 11 '16 at 15:09
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    The only "subtlety" in Smullyan's argument is not about the computations... but about the way he shows that $I_{n+1}=2 \times I_n$. – Mauro ALLEGRANZA Dec 11 '16 at 15:14
  • @MauroALLEGRANZA So set A with n elements would have $2^n$ subsets; I in that Solution, is it taken just a first letter from 'Integer', and S from 'Subset' perhaps, then where does that k come from? As he also mentions x there, too, are x and k the same? – Aili J. Dec 11 '16 at 18:51
  • "the $2k$ sets..." is not a "name"; it means, that : if $I_n$ has $k$ subsets, then $I_{n+1}$ has the double number of subsets, i.e. $2k$. This is the first part of the solution; the second part is to evaluate $k$. He starts from $I_1$, i.e. with $n=1$ ($n$ is the number of the elements of $I_n$) and shows that the number of subsets of $I_1$ is $2$. Now he goes on calculating ... – Mauro ALLEGRANZA Dec 11 '16 at 19:21
  • @MauroALLEGRANZA 5."As we saw in the solution to Problem 1, for any positive integer n , there are exactly $2{^n}{^-}{^1}$ sets of positive integers whose highest number is n,..."(page 25) Sorry but I can't see how - as much as I look there is nothing like this said in Solution 1, Could You Please have a look? – Aili J. Dec 12 '16 at 15:44

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