Let $A\neq \emptyset$ be a set. Show that the following are equivalent:
- $|\mathbb{N}| = \aleph_0\leq |A|$
- $|A| = n|A|$ for any $n\geq 2\in \mathbb{N}$
- $|A| = \aleph_0 |A|$
I want to show $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (1).$
$(1)\Rightarrow (2)$. Suppose $(1)$ holds. Then there is an injection $f : \mathbb{N}\to A.$ It suffices to show that $|A| = |A| + |A| = |A\times 2| = |A\times \{0\}\cup A\times \{1\}|$, where $2 = \{0,1\},$ since an inductive argument can be applied to give the required result. So we need to define a bijection $f : A \to A\times 2.$ Let $\mathcal{F} := \{(X,f) : X\subseteq A, f : X\to X\times 2\text{ is a bijection}\}.$ Observe that $\mathcal{F}$ is nonempty because $A$ contains a denumerable subset $X$, and thus there exists a bijection $f : X\to X\times 2.$ Order the elements of $\mathcal{F}$ by extension. That is, let $(X_1, f_1) \leq (X_2, f_2)$ if $X_1\subseteq X_2$ and $f_2\vert_{X_1} = f_1$. Let $C = \{(X_i, f_i)\}_{i\in I}$ be a chain in $\mathcal{F}.$ Let $X = \cup_{i\in I} X_i$ and $f : X\to X\times 2$ be defined so that $f(x) = f_i(x),$ where $i\in I$ is so that $x\in X_i.$ Then note that $(X,f)$ is an upper bound for $C,$ and is in $\mathcal{F}.$ So by Zorn's Lemma, $\mathcal{F}$ has a maximal element, say $(Y, g).$ Suppose $|Y| < |A|.$ Then $|A\backslash Y| = |A| - |Y| = |A|.$ So $A\backslash Y$ has a denumerable subset $Z.$ Since $Z$ is denumerable, there is a bijection $h : Z\to Z\times 2.$ Then let $i : Y\cup Z\to (Y\cup Z)\times 2, i(w) = \begin{cases}g(w),&\text{ if $w \in Y$}\\ h(w),&\text{ otherwise }\end{cases}$ Then $(Y,g) < (Y\cup Z, i),$ a contradiction. So $|Y| = |A|.$ Hence $|A| = |Y| = |Y\times 2| = |Y| + |Y| = |A| + |A|,$ as required.
$(2)\Rightarrow (3).$ One can find a bijection between $\{1,\cdots, n\}\times A$ and $\{1,\cdots, n+1\}\times A$ for any $n\geq 1$, since $n|A| = (n+1)|A|$ (indeed $n|A| = |A|$ so $(n+1)|A| = n|A| + |A| = |A| + |A| = |A|$). Hence, one can find a bijection between $\{1\}\times A$ and $\mathbb{N}\times A,$ which implies $|A| = |\{1\}\times A| = |\mathbb{N}\times A| = \aleph_0 |A|.$
$(3)\Rightarrow (1).$ If $|A| = \aleph_0 |A|,$ then there is a bijection $f : A \to \mathbb{N}\times A.$ Fix $a \in A$ and let $h : \mathbb{N}\to A, h(b) = f^{-1} (b,a).$ Observe that $h$ is injective because if $b\neq d,$ then since $f^{-1}$ is injective, $h(b) = f^{-1}(b,a) \neq f^{-1}(d, a) = h(d).$ Thus, $\aleph_0 \leq |A|.$
Is this incorrect? Especially $(2)\Rightarrow (3)$ and $(3)\Rightarrow (1).$ Also, WHY is $f$ well-defined in $(1)$? If I choose $i\in I$ randomly, isn't it possible that $x\neq y $ but $x\in X_i$ and $y \in X_j, i\neq j$ and $f_i(x) = f_j(y)$?
(1)$\implies$(2): The idea is very nice that we could in theory stop at any subset of cardinality $|A|$, however, I think, when you stated $|Y|<|A|\implies |A\setminus Y|=|A|$, you implicitly used the statement to be proved: $|Y|+|Y|=|Y|$ for infinite $|Y|$.
Instead, I think, we should prove that $Y=A$: suppose $f:Y\to Y\times 2$ is a bijection and $x\in A\setminus Y$, then we can shift one of the columns in a denumerable subset $Y_1$ of $Y$ in order to obtain a bijection $\,Y\cup\{x\}\,\longrightarrow \,(Y\cup\{x\})\times 2$.
(2)$\implies$(3): It's not correct.
By $2|A|=|A|$, we can cut $A$ into half: there are disjoint $A_0$ and $B_0$ such that $|A_0|=|B_0|=|A|$ and $A=A_0\cup B_0$. This can be done with $B_0$ again, yielding $A_1$ and $B_1$, and then so on, cut $B_i$ to $A_{i+1},\ B_{i+1}$.
Fix bijections $h_n:A\to A_n$, then map $(n,a)\in\Bbb N\times A$ to $h_n(a)$. This is a bijection to $\bigcup_nA_n\subseteq A$.
(3)$\implies$(1): It's correct.