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I have an infinite set $X$, I need to take a proper (countable or not) infinite subset of $X$ so there are theorems that guarantee its existence but it's not clear what set of axiom I need to take, maybe ZFC. Besides, how do I know the complement of that subset is also infinite?

Bernard
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tobiasbriones
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    In general, you cannot do this without some use of the axiom of choice. The counterexamples are called amorphous sets. – Andrés E. Caicedo May 10 '18 at 00:08
  • In ZF we can show that AC (the Axiom of Choice) implies Dependent Choice (DC).. .In ZF+DC) we show that an infinite set $X$ has a countably infinite subset. That is, there exists an injection $f:\Bbb N\to X .$ Then let $Y={f(2n):n\in \Bbb N} .$ Observe that $X$ \ $Y$ is also infinite as it has an infinite subset ${f(2n-1):n\in \Bbb N}.$...... But DC is not a theorem of ZF (unless $1=0$ is too).... A set with a countably infinite subset is also called Dedekind-infinite.... – DanielWainfleet May 10 '18 at 05:00
  • @AndrésE.Caicedo. Is it consistent with ZF that a (Russell) Sock Set can be amorphous? That is, if $F:\Bbb N\to V$ where each $F(n)$ is a $2$-element set, can $\cup F''\Bbb N$ be amorphous? – DanielWainfleet May 10 '18 at 05:05
  • can ∪F′′N be amorphous?.... Only if it is finite. Otherwise define $G(n) = F(n) \setminus \bigcup_{i=0}^{n-1} F(i)$. The non-empty G(n) form a countably-infinite family of disjoint sets and so can be split into two families with disjoint, infinite unions. – David Hartley May 10 '18 at 18:32

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