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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 $X.$

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    $\begingroup$ 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 at 10:47
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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.

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