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 $X.$

  • 1
    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Jan 20 '19 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.