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Consider a continuous bijection $X\to Y$ such that $X$ is homeomorphic to a separable infinite dimensional Hilbert space. I wonder what can be said about topological properties of $Y$.

To exclude silly pathologies (e.g. $Y$ is $X$ with trivial topology), let's suppose that $Y$ is separable and metrizable.

Motivation. Suppose you study a space that appears naturally, and you manage to bijectively parametrize with a continuous parameter that takes values is a nice parameter space (such as $l^2$). If you discover that the parametrization isn't a homeomorphism, you may wonder if you have proved anything at all.

Remarks (to the original question):

  1. $Y$ is path-connected; in fact, no countable (or finite) subset can separate $Y$. (Here is why: the preimage of a countable subset of $Y$ is a countable subset of $X$. Then it is known that the complement in $l^2$ of a countable subset is homeomorphic to $l^2$, and hence the image of the complement in $Y$ is path-connected.)

  2. $Y$ can be homeomorphic to $X$ even when the bijection is not a homeomorphism. In fact, there is a result of Savkin [A criterion for Banach manifolds with a separable model to be finite-dimensional,Ukrainian Mathematical Journal volume 46, pages 1210-1214 (1994)] that any infinite dimensional Banach manifold admits a continuous self-bijection that is not a homeomorphism. See also a recent paper by Creswell in Monthly.

  3. The same question is interesting when $X$ is $\mathbb R^n$. (I briefly thought that the answer in this case follows from the invariance of domain, and Bill kindly corrected me in comments).

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    $\begingroup$ I do not understand (3). The real line bijects continuously onto the figure 6 minus its top point. $\endgroup$
    – Bill Johnson
    Commented Oct 7, 2011 at 0:37
  • $\begingroup$ @Bill: thanks! I obviously misapplied invariance of domain. $\endgroup$
    – Igor Belegradek
    Commented Oct 7, 2011 at 1:01
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    $\begingroup$ A continuous bijection $X$ to $Y$ is just a coarsening of the topology of $X$. There are a lot of coarsenings of that topology. Metrizable coarsenings just correspond to metrics that are continuous in both variables. Obviously, the 6-thing, in which far away points become very close, can happen. I'm actually shocked by my inability to find other examples of bad behavior. The only reason I can imagine that there wouldn't be anything else is if every continuous metric on a compact metrizable space gives the same topology. $\endgroup$
    – Will Sawin
    Commented Oct 7, 2011 at 3:09
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    $\begingroup$ In an infinite-dimensional Hilbert space, since it's not even locally compact, you can do all sorts of weird stuff. Choose a sequence $a_1,a_n,..$ with no limit points and decide on a point $a$ you want it to converge to. Put a bridge of length $1/2^n$ from $a_n$ to $a$, and $a$ becomes the limit of that sequence. You can do this a lot of times, as long as you make sure that you never reduce the distance between two points to $0$. I don't think the separability is necessary. The image of a countable dense subset of $X$ is a countable dense subset of $Y$. $\endgroup$
    – Will Sawin
    Commented Oct 7, 2011 at 3:19
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    $\begingroup$ The bridge isn't actually made of points, it just reduces the distance. $\endgroup$
    – Will Sawin
    Commented Oct 8, 2011 at 4:23

2 Answers 2

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I think I now have a complete answer. Since:

$Y$ is just $X$ with a coarser topology. Metrizable topologies are defined entirely by their convergent sequences. Metrizable topologies are Hausdorff and so convergent topologies can't have subsequences that converge to something else.

The only thing you can do is add new limits to divergent sequences. Thus, $Y$ is just $X$ where some of the divergent sequences (in $\mathbb R^n$, sequences going off to $\infty$, but in a Hilbert space, lots of things) now have limits.

Is that specific enough?

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  • $\begingroup$ @Will Sawin: thanks for thinking about this, but aren't you just restating my question? I wish to have some examples of spaces that can (or cannot) appear as $Y$, and I am not sure how to use your answer for the purpose. $\endgroup$
    – Igor Belegradek
    Commented Oct 7, 2011 at 12:29
  • $\begingroup$ It doesn't seem like there's much more to say. No traditional topologies on vector spaces that I am aware of are coarser than the standard one and metrizable. For $\mathbb R^n$, the only thing you can do is 6: glue paths to infinity to points nearby. Locally (compactly), the topology is preserved. In Hilbert space, you can do all kinds of crazy stuff, but the only thing that's both consequential and admits comprehensible description (that I am aware of) is the 6 thing again. $\endgroup$
    – Will Sawin
    Commented Oct 8, 2011 at 4:32
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Note that $X$ is homeomorphic to $\mathbb{R}^\omega$ and therefore it contains a copy $C \subseteq X$ of the Hilbert cube $[0,1]^\omega$. Since the Hilbert cube is compact, the given bijection restricted to $C$ is a homeomorphism onto its image, so $Y$ must also contain a copy of the Hilbert cube. From here, $Y$ must be Universal for the class of separable metrizable spaces.

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  • $\begingroup$ This is a nice observation, thanks. $\endgroup$
    – Igor Belegradek
    Commented Feb 13, 2013 at 17:16
  • $\begingroup$ By the way, in your answer $Y$ must be Hausdorff. $\endgroup$
    – Igor Belegradek
    Commented Jun 20, 2013 at 19:03

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