Set theory cardinality question

  • Thread starter klabatiba
  • Start date
  • #1
1
0
Can anyone please give a really explicit proof (omitting no steps) and with as simple words as possible that any infinite set can be writtern as the union of disjoint countable sets?

Thank you.
 

Answers and Replies

  • #2
147
0
How about something like this:

Every infinite set X has a countable subset (this is a theorem of ZFC). We'll construct a function f from an initial set of ordinals to disjoint countably infinite subsets of X such that X its range. Let f(0) be any countably infinite subset of [itex]X = X_0[/itex]. Then define [itex]X_{i+1} = X_i \setminus f(i)[/itex], [itex]X_i = \displaystyle{ \cap_{j<i} X_j}[/itex] for a limit ordinal i and let [itex]f(i)[/itex] is a countably infinite subset of [itex]X_i[/itex] (the possibility of making these choices collectively relies on the Axiom of Choice). The sequence must eventually end with a null set - otherwise, the set X would be larger than all cardinals.
 
  • #3
1,030
4
Well, any set is a union of singletons containing exactly one element from the set. A union of all these singletons (which are finite thus countable) gives you that set.
 
  • #4
286
0
I mention only that however the countable sets are constructed, the union must consist of an uncountable number of those sets.

--Elucidus
 
  • #5
147
0
Well, obviously, countable sets cannot be expressed as an uncountable union of disjoint countable sets (assuming by countable you mean countably infinite, or at least non-empty).
 

Related Threads on Set theory cardinality question

  • Last Post
Replies
12
Views
5K
Replies
1
Views
1K
Replies
2
Views
873
Replies
3
Views
663
M
  • Last Post
Replies
1
Views
988
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
3
Views
2K
Top