Let each $(A_i,T_i) = (\{0,2\}, T_{discrete})$ and define $\phi : \prod (A_i, T_i) \rightarrow [0,1]$ with $\phi (<a_1, a_2, ...>) = \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}}$.
If we let $W = \{a_1\} \times \{a_2\} \times ... \{a_N\} \times A_{N+1} \times A_{N+2} \times ...$
What is $\phi (W)$?
It seems to be $\frac {a_1}{2^2} +\frac {a_2}{2^3} + ... + \frac {a_N}{2^{N+1}} + \frac{\{0,2\}}{2^{N+2}}$ + ... ? but this doesn't make sense.
HINT: You’re right: as you’ve written it, it doesn’t really make sense. $W$ contains infinitely many different points of the product, each of which is a sequence of $0$s and $2$s, and $\varphi[W]$ contains the images under $\varphi$ of all of these sequences. Thus,
$$\varphi[W]=\left\{\frac{a_1}{2^2}+\frac{a_2}{2^3}+\ldots+\frac{a_N}{2^{N+1}}+\sum_{n\ge N+1}\frac{a_n}{2^{n+1}}:a_n\in\{0,2\}\text{ for each }n\ge N+1\right\}\;.$$
Now notice that
$$\frac{a^n}{2^{n+1}}=\begin{cases} \dfrac0{2^{n+1}}=\dfrac0{2^n},&\text{if }a_n=0\\ \dfrac2{2^{n+1}}=\dfrac1{2^n},&\text{if }a_n=2\;, \end{cases}$$
so if we set $b_n=\frac{a_n}2$ for $n\ge 1$, we can rewrite $\varphi[W]$:
$$\varphi[W]=\left\{\frac{b_1}{2^1}+\frac{b_2}{2^2}+\ldots+\frac{b_N}{2^N}+\sum_{n\ge N+1}\frac{b_n}{2^n}:b_n\in\{0,1\}\text{ for each }n\ge N+1\right\}\;.$$
At this point it’s very helpful to think of the elements of $\varphi[W]$ interms of their binary representations: for any specific choice of values of the $b_n\in\{0,1\}$, the sum of the infinite series
$$\frac{b_1}{2^1}+\frac{b_2}{2^2}+\ldots+\frac{b_N}{2^N}+\sum_{n\ge N+1}\frac{b_n}{2^n}\tag{1}$$
is simply the real number whose binary representation is
$${0.b_1b_2\ldots b_Nb_{N+1}\ldots\,}_{\text{two}}\;:\tag{2}$$
by definition the binary expansion $(2)$ means the sum of the series $(1)$. Thus, $\varphi[W]$ is the set of all real numbers that have binary expansions starting $0.b_1b_2\ldots b_N$.
Once you’ve answered those questions, you should be able to describe $\varphi[W]$ in very simple terms.