Why $\epsilon \aleph_0 =\epsilon$?

Question

Just a quick question. In the proof of Proposition. 4.14. in Conway's Functional Analysis, it states without any proof that if $E$ is any infinite set with cardinality $\epsilon$ then $\epsilon \aleph_0 =\epsilon$ holds, I just know the proof for special case $\epsilon = \aleph_0$ and not for $\epsilon= \aleph_1, ...$ and I know very little set theory. Please help!

Answer 1

(I also don't know much set theory, so this may not be right.)

Once we know the fact that $\aleph_0^2 = \aleph_0$, we can generalize it to $\epsilon\aleph_0 = \epsilon$. Let $E$ be a set with cardinality $\epsilon$. We will show $|E| = |E \times \mathbb{N}|$. Clearly, $|E| \leq |E \times \mathbb{N}|$, so it remains to show the other direction.

First, divide $E$ into a family $\{E_i\}_{i \in I}$ where each $E_i$ is a countably infinite subset of $E$. Then for each $E_i$, there is an injective function $\phi_i : E_i \times \mathbb{N} \to E_i$. We can construct an injective function $\phi : E \times \mathbb{N} \to E$ where

$$ \phi(e, n) = \phi_i(e, n) \quad \text{for $e \in E_i$}. $$

Therefore, $|E \times \mathbb{N}| \leq |E|$, and so we conclude $\epsilon\aleph_0 = \epsilon$.

Answer 2

Assume $\epsilon$ is an initial ordinal. Each of its elements is equal to $\lambda + n$ for some unique limit ordinal (or $0$) $\lambda$ and $n \in \mathbb N$. Let $\psi$ be a bijection $\mathbb N \times \mathbb N\to \mathbb N $. Then mapping $(\lambda + n, m) \mapsto \lambda + \psi(n,m)$ gives a bijection $\epsilon \times \mathbb N \to \epsilon$.

If you assume the Axiom of Choice, all infinite sets are equal in cardinality to some initial ordinal and so the result holds for all cardinals. Without choice it may fail, for example if $\epsilon$ is Dedekind finite then $\epsilon \aleph_0$ will be Dedekind infinite and so cannot equal