4
$\begingroup$

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 $R$?

$\endgroup$
  • 2
    $\begingroup$ 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 $X$. $\endgroup$ – erz Mar 30 '18 at 1:06
  • $\begingroup$ Now I wonder if there is an intrinsic characterization of such spaces. $\endgroup$ – erz Mar 30 '18 at 1:09
  • 1
    $\begingroup$ 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. $\endgroup$ – erz Mar 30 '18 at 1:24
  • $\begingroup$ math.stackexchange.com/questions/62820/… $\endgroup$ – erz Mar 30 '18 at 1:24
1
$\begingroup$

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].
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.