We define an embedding of the set of prim numbers into the Cantor set as follows:


First we recall that the cantor set $\mathcal{C}$ is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $ since the latter is a compact metrizable space without any isolated point. So according to [topological characterization of the Cantor set](https://en.wikipedia.org/wiki/Cantor_space) the classical Cantor set is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $.


The space of prime numbers is denoted by $\mathcal{Prime}$.


We define the embedding $\mathcal{E}:\mathcal{P}\to (\mathbb{Z}/10\mathbb{Z})^\omega $ as follows:


$$\mathcal{E}(p)=(a\_1,a\_2,\ldots,a\_n,\ldots)$$


where the decimal expansion of $\sqrt{p}=b\_nb\_{n-1}\ldots b\_{1}b\_0/a\_1a\_2\ldots a\_n\ldots$


So in this way we may consider the space of prime numbers $\mathcal{Prime}$ as a subspace of the Cantor set $\mathcal{C}$.



> 
> Is $\mathcal{Prime}$ a compact set? Is it an open subset of $\mathcal{C}$?
> 
> 
> 


What would be a number theoretical interpretations for these topological questions?



The set of prime numbers as a subspace of the Cantor set
Your set is a countable dense subset of $(\mathbb{Z}/10\mathbb{Z})^\omega$; cf. Lucia's response [here](https://mathoverflow.net/questions/292044/distribution-of-square-roots-mod-1). Hence it is neither open, nor compact.


