# Set theory cardinality question

1. Sep 10, 2009

### klabatiba

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.

2. Sep 10, 2009

### Preno

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 $X = X_0$. Then define $X_{i+1} = X_i \setminus f(i)$, $X_i = \displaystyle{ \cap_{j<i} X_j}$ for a limit ordinal i and let $f(i)$ is a countably infinite subset of $X_i$ (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. Sep 10, 2009

### Dragonfall

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. Sep 10, 2009

### Elucidus

I mention only that however the countable sets are constructed, the union must consist of an uncountable number of those sets.

--Elucidus

5. Sep 11, 2009

### Preno

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).