Eventually-prime decimal expansions Let $w$ be a right-infinite word over the alphabet $A = \{ 0, 1, \dots, 9\}$, with a distinguished decimal point after at most finitely many symbols from the left (i.e. $w$ is in $A^\ast . A^\omega$). For example, we can consider $w_{\sqrt{2}} = 1.41421356237\dots$, the decimal expansion of $\sqrt{2}$, or $w_p = 0.235711131723\dots$, where $w_p$ is the concatenation of all the primes.
This second word $w_p$ has a curious property, namely that it is a subword of $0.P^\omega$, where $P$ is the language $\{ 2, 3, 5, 7, 11, \dots\}$ of prime strings. In other words, we can always factor the word following the decimal into primes (in at least one way!).
Let us say that the right-infinite word $w = w'.w''$ with $w' \in A^\ast$ and $w'' \in A^\omega$ has property $\mathfrak{P}$ if $w'' \in P^\omega$. Some words do not have this property: for example, the word $0.444444\dots$ does not have property $\mathfrak{P}$.
(1) Does $w_{\sqrt{2}}$ have property $\mathfrak{P}$? Are there some other "canonical" constants $c$ (e.g. $c= \pi, e, \sqrt{3}, \zeta(2), \dots$) which give rise to words $w_c$ with property $\mathfrak{P}$?
A way to think about the question for $w_{\sqrt{2}}$ is as follows: following the decimal point, we read the right-infinite word $4142135623730950 \dots$. Starting from the left, we see $41$, which is prime; so if the word $421356237 \dots$ is in $P^\omega$ then $w_{\sqrt{2}}$ has property $\mathfrak{P}$. On the other hand, $4142135623$ is also prime (as can be checked), so if $730950 \dots$ is in $P^\omega$, we can also conclude that $w_{\sqrt{2}}$ has property $\mathfrak{P}$. There is thus some non-determinism in the factorisation chosen.
(2) Can anything be said about the density (appropriately defined) of decimal words with property $\mathfrak{P}$ in the interval $[0,1]$?
There are some subtleties about identifying numbers and decimals, e.g. $0.89999\ldots = 0.9$, but as $999\dots 999$ is never prime this is hopefully not too great of an issue (and can probably be formalised away). Of course, the question can also be asked about binary right-infinite strings, etc., which may be easier.
 A: Let me answer the second of your two questions. (I don't expect that the answer to your first question is within the reach of today's mathematics.)
Let $S$ be the set of real numbers in $[0,1]$ with property $\mathfrak{P}$.
$S$ is indeed dense in $[\dfrac{1}{10},1]$. If the first digit after the decimal point is $0$, then the string is not in $P^\omega$, so we start the interval at $\dfrac{1}{10}$ and not at $0$.
You just need to show that given $r$ strictly between $0$ and $1$ such that $r$ has finite decimal expansion $a_1\ldots a_n$ and zeroes after,
for an arbitrarily large $k$ you can find a prime such that the first $n$ digits are $a_1\ldots a_n$ and the next $k$ digits are zeroes.
Let $q$ be the $n$-digit whole number of which the digits are $a_1\ldots a_n$.
Notice that for fixed $k$,
if $N$ is large enough, there is at least one prime strictly between $10^N q$ and $10^N q+ 10^{N-k}-1$. You can prove this using the theorem that the number of primes up to $x$ is asymptotic to $\dfrac{x}{\ln(x)}$.
$\\$
