My first try was using Cantor style diagonal argument. Let's take Si ⊆ℕ be an infinite subset of natural number. Then Si can be represented as a binary sequence. Let's take a collection of subset in a way that Si has 1s in all the location where Si-1 has 1, but Si has a 1 in at least one location where Si-1 has 0. This way, Si-1 ⊆ Si. Let's take the following example to demonstrate:
S1 = 0 1 0 1 0 0 1 0...
S2 = 1 1 0 1 0 0 1 0...
S3 = 1 1 0 1 0 1 1 0...
.
.
.
Now, following Cantor's diagonal argument we can construct a new subset Sn like this:Let's take the first position of the first subset and flip it - That is, take 1 if it was 0, or take 0 if it was 1. Then take this flipped bit as the first element of An. Then go to the second element of the second set and similarly flip that and take that as the second element of An and so on. This way, a totally new subset will be formed which is definitely not in the given list of subsets. But does this new subset conform to the ordering relation asked by the problem? Seems like the very diagonal argument will ensure that it will not. Even if I don't flip the bits, how do I show if this news set is a subset or superset of any of the existing one? Also, will it be a new set if I don't flip the bits?
I thought of another way. Let's assume we got a collection of subsets of Si ⊆ℕ such that for any Si and Sj, either Si ⊆ Sj or Sj ⊆ Si. Such collection of subsets definitely do exists. For example if we take ℕ and sequencially drop the smallest element from them, the resulting collection of subsets {0, 1, 2, 3 ...}, {1, 2, 3 ...}, {2, 3 ...}, ... will be such a collection. But it will also be countably infinite. So, how do I prove or demonstrate that there do exist such a collection that is uncountable?
I've encountered this problem while trying to self-study real analysis. The specific book in question is: "The Real Numbers" by John Stillwell. I'm not a mathematics student. This is just my personal quest. Please help.
