Showing $\prod_{n < \omega} n = 2^{\aleph_0}$

Question

I have to show that $\prod_{n < \omega} n = 2^{\aleph_0}$. I'm having trouble getting started. I know that $2^{\aleph_0}$ is the set of binary sequences, or the space of functions from $\mathbb{N}$ to $\{0, 1\}$, but I'm having trouble parsing $\prod_{n < \omega} n$. My best guess is that it's the infinite Cartesian product $$0 \times 1 \times 2 \times 3 \times \dots$$ but because $0$ is $\emptyset$, then so is this product. Alternatively, $$\prod_{n < \omega} n = \{f : \omega \rightarrow \bigcup n | (\forall n) f(n) \in n\}$$ Thus has the same problem, though, that this would be an empty space of functions because none of them can map $0$ anywhere.

Answer 1

It's certainly true that $\prod_{n < \omega} n = 0$, so I presume what was meant is $\prod_{0 < n < \omega} n$, in which case this is the set of functions $f : \omega \setminus \{0\} \to \omega \setminus \{0\}$ with $f(n) \in n$ for all $0 < n < \omega$, i.e. the set of regressive functions $\omega \setminus \{0\} \to \omega \setminus \{0\}$. (This has the same cardinality as the set of regressive functions $\omega \to \omega$.)

More generally, if $A$ is an indexing set and for each $\alpha \in A$ there is a set $X_{\alpha}$, then $\prod_{\alpha \in A} X_{\alpha}$ is the set of functions $f : A \to \bigcup_{\alpha \in A} X_{\alpha}$ such that $f(\alpha) \in X_{\alpha}$ for all $\alpha \in A$.

So you need to prove that there are $2^{\aleph_0}$-many regressive functions $\omega \to \omega$.

One method using cardinal arithmetic as as follows:

• To prove $2^{\aleph_0} \le \prod_{0<n<\omega} n$, use the fact that $2 \le n$ for all $n \ge 2$ (duh), so $$\prod_{0 < n < \omega} n \quad = \quad 1 \times \prod_{1<n<\omega} n \quad \le \quad \cdots$$
• To prove $\prod_{0<n<\omega} n \le 2^{\aleph_0}$, you can do a similar trick, namely to note that $n \le 2^n$, and so $$\prod_{0<n<\omega} n \le \prod_{0<n<\omega} 2^n = 2^{\sum_{0<n<\omega} n} = \cdots$$

You'll need to prove both the inequalities I stated and complete the $\cdots$ bits.

You could also prove this using cleverly chosen set inclusions.

Answer 2

Hint. The product is bounded below by $\prod 2$ and above by $\prod \aleph_0$.