Can the list of finite subsets of the positive integers be put in a 1-to-1 correspondence to the set of positive integers?

Question

If so, can you demonstrate it with a specific example which shows the beginning of a listing of such a correspondence?

Have you heard of Hilbert's Hotel? https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel — Mark Bennet, Dec 03 '22 at 17:45
Does this answer your question? Show that the set of all finite subsets of $\mathbb{N}$ is countable. — José Carlos Santos, Dec 03 '22 at 17:46
The list of finite subsets is itself infinite. ANd yes it is countable. And here is a way to do it: map a subset $S$ to the positive integer $\sum_{s\in S} 2^{s-1}$. SO that way ${1,2,5,7} \to 2^0 + 2^1+ 2^4 +2^6= 1 + 2+16 + 32=51$. I'll leave it to you to see why that is one to one. I'm assuming you are not allowing $\emptyset$ and $0$ but it can be easily modified for those cases. — fleablood, Dec 03 '22 at 17:52
One thing I find kind of interesting is that you can use Cantors diagonal argument to prove that list of finite subsets isn't finite. (List a finite list of subset and take a set by modifying the $k$th term of the $k$ subset). You end up with a finite subset not on a list so no finite list contains the all. I can also you Cantors argument to prove the powerset of ${1,2....,7}$ must have cardinality of more than $7$.... but that'd be kind of silly. — fleablood, Dec 03 '22 at 17:57
@Jose Carlos Santos -- That link would potentially answer the first part of my question. — Olive Stemforn, Dec 04 '22 at 01:55

score 0 · Answer 1 · answered Dec 12 '22 at 17:18

Here is one example of how to order a list of the finite subsets of the positive integers:

{1},

{2},

{3}, {1, 2},

{4}, {1, 3},

{5}, {1, 4}, {2, 3},

{6}, {1, 5}, {2, 4}, {1, 2, 3},

{7}, {1, 6}, {2, 5}, {3, 4}, {1, 2, 4},

{8}, {1, 7}, {2, 6}, {3, 5}, {1, 2, 5}, {1, 3, 4},

. . .

1 Answers1