# Approaching a space with a ray

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 $R$?

• Let $X$ be a Peano continuum, i.e. there is a continuous surjection $p:[-1,1]\to X$. Define $r:(0,1]\to X\times [0,1]$ by $r(t)=(p(\sin(1/t)),t)$. It looks like this is a homeomorphism from a ray onto its image $R$. Any point of $X$ is approximated by the the points of $R$, and so $Y=R\cup X\times\{0\}$ meets your conditions ($R$ is open because $X\times\{0\}$ is closed). Also, for any $X'\subset X$ consider $Y'=R\cup X'\times\{0\}$. Thus, any space, which is a subset of a locally connected metrizable compact can serve as $X$. – erz Mar 30 '18 at 1:06
• Now I wonder if there is an intrinsic characterization of such spaces. – erz Mar 30 '18 at 1:09
• Ok, it says in the link below, that any separable metric space is a subset of the Hilbert Sube, which is a Peano continuum. Hence, $X$ can be any separable metic space. – erz Mar 30 '18 at 1:24
• math.stackexchange.com/questions/62820/… – erz Mar 30 '18 at 1:24

## 1 Answer

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].