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.
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 $\dfrac{x}{\ln(x)}$.
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