# Partition of an infinite set into two infinite sets

I wanted to try and prove this statement which looks seemingly true.

An infinite set $$X$$ can be partitioned in such a way that $$X = X_1 \cup X_2$$ where $$X_1$$ and $$X_2$$ are infinite subsets of $$X.$$

Attempt :

case (i) If $$X$$ is countably infinite, then we can list the elements of $$X$$ as $$\{x_1,x_2,x_3,\dots\}.$$ If we choose $$X_1$$ to be indexed by the odd natural numbers and $$X_2$$ to be indexed by the even natural numbers, we are done.

case (ii) If $$X$$ is uncountably infinite. Choose $$X_1$$ to be the countable infinite subset of $$X.$$ Then let $$X_2 := X_1^\mathsf{c}.$$ This shall ensure that $$X_2$$ is uncountable. $$\space\space\blacksquare$$

But I think I'll have to be a bit clear for case (ii). How do I show that given an uncountable set $$X,$$ there always exists a countable infinite subset $$X_1$$ of $$X.$$

• Every infinite set contains a countably infinite set (if we assume some axiom of choice). An uncountable set is in particular infinite. Also see this thread etc. Jan 20, 2019 at 10:47

Case 2. Well order X and take the first $$\omega_0$$ elements.
Or recursively construct a denumerable set. Pick any element a$$_1$$ from X. Having picked S = { $$a_1,.. a_n$$ }, pick any
a$$_{n+1}$$ from uncountable X - S.