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I am interested in topological spaces such that whenever the space embeds into the Hilbert cube, the image of the embedding has a path-connected complement.

Any finite dimensional space has this property by an argument based on Alexander duality in a finite dimensional approximation of the Hilbert cube, see e.g. Lemma 2.1 in "Characterization of finite-dimensional 𝑍-sets" by Kroonenberg [Proc. Amer. Math. Soc. 43 (1974), 421-427].

Are there other examples?

UPDATE:

  1. First of all, a correction: the above mentioned Lemma 2.1 needs an assumption that the image of the embedding is closed. Without this assumption I can only show that the complement of a finite dimensional subset of a Hilbert cube is connected. (The proof is the same where instead of Alexander duality one uses the result that codimension two subspaces of $\mathbb R^n$ have connected complements).

  2. The simplest example of a subset of the Hilbert cube with path-connected complement is that of deficiency $\ge 2$, where deficiency of a subset is the number of coordinate projections mapping the subset to a point. I could not find any study of subsets of deficiency $\ge 2$. Rather, people studied closed subsets of infinite deficiency, or their images under ambient homeomorphisms which are also known as the $Z$-sets. So one could ask for conditions on a space such that any embedding in the Hilbert cube is a $Z$-set. The definitive answer is given be Lenaburg in "Absolute $Z$-sets": any space with this property is countable! Thus we reach a sad conclusion: looking at closed subsets of infinite deficiency does not give any new examples of spaces that always separate the Hilbert cube no matter how we embed them.

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By the same argument $\sqcup_n \mathbb R^n$ also has this property. If you want the space to be connected you could take a wedge instead of a disjoint union. –  Ryan Budney Jul 1 '13 at 2:41
    
Ryan, the argument in the linked paper seems to fail right before one invokes duality. Did you have in mind some modification? –  Igor Belegradek Jul 1 '13 at 3:00
    
@Tom, you surely meant to ask something else. Any finite dimensional compact subset of the Hilbert cube satisfies your condition; it does not separate by the Alexander duality, and it cannot contain an infinite dimensional subspace, such as the Hilbert cube. –  Igor Belegradek Jul 1 '13 at 18:25
    
Also any separable metrizable space embeds into the Hilbert cube (by Urysohn's embedding). –  Igor Belegradek Jul 1 '13 at 19:30
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Yes, Igor, there was an erroneous "not". I've deleted the comment. Here's what I meant: I cannot think, off the top of my head, of any subset $X\subset Q$ of the Hilbert cube that separates $Q$ and that does not have a subset homeomorphic to $Q$. But I have little experience with these things and have not thought very hard. –  Tom Goodwillie Jul 2 '13 at 19:47
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