Cardinality if set $A$ equals $n|A|$ implies $n|A| = \aleph_0|A|$

Question

Suppose $|A| = n|A|$, for any $n\in \mathbb{N}$. Then $|A|=\aleph_0 |A|$. How does one prove this? I'm feeling lost. Would appreciate some hints. Is Zorn's lemma useful here?

Intuitively, I would think that since $n$ is arbitrarily large, and $|A|=n|A|$, $n|A|=\aleph_0|A|$. But this doesn't seem quite rigorous.

Answer 1

Here's a proof without the axiom of choice.

We assume that $|A|=2|A|=|\{0,1\}\times A|.$

I.e., there is an injective map $e:\{0,1\}\times A\to A.$

For $a\in A$ define $f(a)=e(0,a)$ and $g(a)=e(1,a)$;
then $f:A\to A$ and $g:A\to A$ are injective maps with disjoint ranges.

For $n\in\mathbb N=\{0,1,2,\dots\}$ and $a\in A$ define $h(n,a)=f^ng(a),$
i.e., $h(0,a)=g(a),\ h(1,a)=f(g(a)),\ h(2,a)=f(f(g(a))),$ etc.

It's easy to see that $h:\mathbb N\times A\to A$ is an injective map, showing that $|\mathbb N\times A|\le|A|.$ Combining this with the obvious inequality $|A|\le|\mathbb N\times A|$ and the Cantor–Bernstein theorem, we get
$|A|=|\mathbb N\times A|=\aleph_0\cdot|A|.$

Answer 2

Assume $|A|$ is finite. Then, if $|A|$=$n|A|$ for each $n$ a natural number, $|A|=0$. Therefore, $|A|=\aleph_0|A|=|\emptyset\times\mathbb{N}|=0$.

Assume otherwise. Let $|A|=\aleph_\alpha$. Then, $\aleph_0\cdot\aleph_\alpha=\mathrm{max}\{\aleph_0,\aleph_\alpha\}=\aleph_\alpha$.

In case you aren't familiar with cardinal arithmetic, try to show that $\aleph_\alpha\cdot\aleph_\alpha$ has a bijection with $\aleph_\alpha$. Then, $\aleph_\alpha\cdot\aleph_\alpha\geq\aleph_0\cdot\aleph_\alpha\geq\aleph_\alpha$ and $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$, meaning $\aleph_\alpha=\aleph_\alpha\cdot\aleph_0$.