Does the multiplication of countably infinite many numbers between $0$ and $1$ equal $0$?

Let $(a_i)$ be a strictly increasing sequence of positive numbers which is bounded above. Such a sequence necessarily converges, call the limit $a$. Now set $x_i = \frac{a_i}{a_{i+1}}$. Then $0 < x_i < 1$ and

$$\prod_{i=1}^nx_i = x_1\dots x_n = \frac{a_1}{a_2}\dots\frac{a_n}{a_{n+1}} = \frac{a_1}{a_{n+1}}.$$

Therefore,

$$\prod_{i = 1}^{\infty}x_i = \lim_{n\to\infty}\prod_{i=1}^n x_i = \lim_{n\to\infty}\frac{a_1}{a_{n+1}} = \frac{a_1}{a}$$

which is non-zero as $a_1 \neq 0$. It follows from this construction that any $p \in (0, 1)$ arises as such an infinite product.

$$\frac{1}{2}=\frac{1}{2^{1/2}}\cdot \frac{1}{2^{1/4}}\cdot \frac{1}{2^{1/8}}\cdot \frac{1}{2^{1/16}}\cdot \frac{1}{2^{1/32}}\cdot \frac{1}{2^{1/64}} \cdots$$

Start with the decreasing sequence $$p_1 = 1 - \frac{1}{4}, \quad p_2 = 1 - \frac{3}{8}, \quad p_3 = 1 - \frac{7}{16}\quad, \quad ...\quad , \quad p_n = 1 - \frac{2^n-1}{2^{n+1}} $$ whose limit equals $\frac{1}{2}$. Then set $$x_1 = p_1, \quad x_2 = p_2/p_1, \quad x_3 = p_3/p_2,\quad \ldots $$ The partial product $\prod_{i=1}^n x_i$ equals $p_n$, and so the limit is $\frac{1}{2}$.