The goal was to find an uncountable set $S$ such that each member of $S$ was a subset of the naturals, and given any two elements $A, B \in S$, either $A \subset B$ or $B \subset A$.
Solution
Consider the rationals in $[0, 1]$ which are countable, say $\{q_1, q_2, \dots\}$
For each real $x \in [0, 1]$ let $S_x = \{i : q_i \le x\}$.
$S = \{S_x : x \in [0, 1]\}$ is a set with that property.
Solution
Consider the rationals in $[0, 1]$ which are countable, say $\{q_1, q_2, \dots\}$
For each real $x \in [0, 1]$ let $S_x = \{i : q_i \le x\}$.
$S = \{S_x : x \in [0, 1]\}$ is a set with that property.
For each natural number n and real number x in (0, 1), map (n, x) to {1,2,...,n}. The resulting sets have duplicates though.
ReplyDeleteI don't think mine is valid.
DeleteYeah, the only sets you have are {1}, {1,2}, {1,2,3}, ... which is countable.
Delete