Sufficient conditions for existence of injection from a metric space $M$ to $\mathbb{R}$ Let $M$ be any metric space. What conditions are required of $M$ for there to exist an injective, continuous function 
$$\varphi \colon M \longrightarrow \mathbb{R}$$
I would like to believe that $M$ just needs to have the same cardinality as $\mathbb{R}$. Would I be correct to think this? If so, how would I prove it?
 A: Such an injective map induced a total order on $M$ in a natural way: $p\le q$ if $\varphi(p)\le \varphi(q)$.  This order has the following property: for any point $p\in M$, the sets $\{x\in M: x<p\}$ and  $\{x\in M: x>p\}$ form a separation of $M\setminus \{p\}$, meaning the closure of either set is disjoint from the other set. Such an order is called a natural order (at least in Whyburn's Analytic Topology, where I saw this definition). Thus, the existence of a natural order is a necessary condition for the existence of $\varphi$. 
Theorem III.6.4 in Analytic Topology gives a sufficient condition: for every compact naturally ordered space there is an order-preserving homeomorphism onto a subset of $\mathbb{R}$. 
Of course compactness isn't necessary. But I doubt there is an intrinsic necessary and sufficient condition for the existence of $\varphi$, because the space $M$ can have very strange topology: it can be the graph of any function (even everywhere discontinuous) from $\mathbb{R}$ into any metric space.
