I know that a quotient space can be thought of as being an open continuous image of a space. Therefore, it would be enough to find some map from $ \beta \mathbb{N}$ open and continuous to $ \beta \mathbb{R} $.
I thought perhaps we could use the Stone-Cech compactification extension property to find this map. In particular one would find a map $ F : \beta \mathbb{N} \rightarrow \mathbb{R}$ that is not only continuous (as guaranteed by the extension property) but open as well. However, I don't see any reason why the map should be open.
I know that $ \mathbb{N} $ and it's complement are closed in $ \mathbb{R}$. I don't know if that can help though.
I think maybe $\mathbb{R}$ can be thought of as a quotient space of $\mathbb{N}$ but i'm not sure that's a way to go either (perhaps by composing said map by the Stone-Cech map for $ \mathbb{N}$?)
Any hint/help would be great. I did look around for something similar. Thanks!
Let $X$ be an arbitrary separable Tychonoff space (in particular, we can consider $X=\Bbb R$) and $n:\Bbb N\to X$ be an arbitrary map with dense image. Let $\beta n:\beta\Bbb N\to\beta X$ be the extension of the map $n$. Then $\beta n(\beta\Bbb N)$ is a compact dense subset of a compact space $\beta X$, so $\beta n(\Bbb N)= \beta X$. Since $\beta n$ is a continuous map of a compact Hausdorff space into a compact Hausdorff space, it is closed, and, therefore, quotient.