Given a separable metric space $X$, what are some ways of forming a new metric space $Y$ such that:


(i) $Y$ contains a ray $R\simeq [0,\infty)$ ($\simeq$ means homeomorphic);


(ii) $R$ is open and dense in $Y$;


(iii) $X\simeq Y\setminus R$.


Moreover, can any non-compact separable metric space play the role of $R$?



Approaching a space with a ray
As @erz observes, if you do such construction for $X$ obtaining $Y$, for any $X' ⊆ X$ you may put $Y' := (Y \setminus X) ∪ X'$, and it still has all three properties. And since you can do this for $X$ being the Hilbert cube, you can to this for any separable metrizable $X$.


On the other hand, it is often additionally required that $Y$ is compact. So you are looking for metrizable compactifications of the ray with $X$ being the remainder (sometimes called *spirals over $X$*). There are many results regardings these.


* In 1932 Waraszkiewicz constucted a family of continuum many such compactifications with $X$ being the circle (such are called just *spirals*) such that they are pairwise incomparable. In this context, two spaces are incomparable when not only they are non-homeomorphic but there is no continuous surjection from one to the other one and vice versa. [Z. Waraszkiewicz, *Une famille indénombrable de continus plans dont aucun n’est l’image d’un autre*, Fund. Math., 18 (1932), pp. 118–137]
* Recently, Pyrih and Vejnar gave simpler proof of the Waraszkiewicz's reuslt. [P. Pyrih and B. Vejnar, *Waraszkiewicz spirals revisited*, Fund. Math., 219 (2012), pp. 97–104]
Essentially, you may assign to a sequence of integers a spiral by the integers coding the winding pattern of the spiral. If the sequences are sufficiently distinct, then the resulting spirals are incomparable.
* These ideas were further generalized in [A. Bartoš, R. Marciňa, P. Pyrih, and B. Vejnar, *Incomparable compactifications of the
ray with Peano continuum as remainder*, Topology Appl., 208 (2016), pp. 93–105.] where $X$ is arbitrary fixed Peano continuum.
* If incomparability is weakened to non-homeomorphism then $X$ can be arbitrary fixed nondegenerate continuum [P. Minc, *$2^{ℵ\_0}$ ways of approaching a continuum with $[1,∞)$*, Topology Appl., 202 (2016),
pp. 47–54]. This cannot be generalized to the incompacarble case since for example all spirals over the pseudo-arc are comparable [A. Illanes, P. Minc, and F. Sturm, *Extending surjections defined on remainders of metric
compactifications of $[0,∞)$*, Houston J. Math., 41 (2015), pp. 1325–1340].


