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Let H be an infinite dimensional and separable Hilbert space. Let C be a closed and connected subset of H containing more than one point. Can C ever be the countable union of closed and totally disconnected subsets of H?

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Let $E_c$ be the complete Erdös space (Erdös, Annals of Math vol 41 1940), defined as the subspace of $\ell^2(\mathbb{N})$ where all coordinates are irrationals. It is polish (separable and completely metrizable) and totally disconnected, but admits a "connectification" namely a (still polish) topology on $E_c\cup\{p\}$ that makes it connected (and of course induces the one on $E_c$). The crucial point is the fact that any nonempty closed and open subset of $E_c$ is unbounded. Then, as in Bill Johnson's answer $E_c$ is the union of the closed and totally disconnected subspaces $\overline{B}(0,n)\cup\{p\}$, $n\geq 1$. It remains to remark that, like any polish space, $E_c\cup\{p\}$ embeds as a closed subset of $H$ (as remarked in Gerald Edgar's answer).

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$C$ is simply a connected complete separable metric space with more than one point. Now the union of countably many closed sets of topological dimension zero must again have dimension zero. But I suppose "totally disconnected" is not quite the same as "zero dimensional" so this is not yet a complete answer.

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  • $\begingroup$ These are the same thing if the set is compact. There exist separable, metrizable, totally disconnected spaces of all dimensions. $\endgroup$ Aug 14, 2010 at 19:58
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Let $C$ be an explosion point space such as the Knaster–Kuratowski fan (http://en.wikipedia.org/wiki/Knaster–Kuratowski_fan) and $p$ the explosion point in $C$.
Set $A_n = \{p\}\cup (C\sim B(p; 1/n))$.

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  • $\begingroup$ Now the K-K fan is not complete (not a $G_\delta$ set in $\mathbb{R}^2$), right? So it cannot be embedded as a closed subset of Hilbert space. $\endgroup$ Aug 15, 2010 at 10:24
  • $\begingroup$ Oops! I missed that condition. Thanks. $\endgroup$ Aug 15, 2010 at 14:16
  • $\begingroup$ Thanks for that example based on the complete Erdos space which answers my question in the affirmative. But I would like to clarify a couple of points. If E is the complete Erdos space and F is the connected metric space in which E is embedded, is F obtained by adjoining a single point p to E? If so, how is F metrized? In other words, how are the distances from p to the other points of F defined? $\endgroup$ Aug 18, 2010 at 19:43
  • $\begingroup$ Yes, $F$ is $E$ with one point added. But exhibiting a complete metric on $F$ is not obvious. At least a metric defining the "right" (connected) topology is implicit in the proof of theorem 1 in this reference : few.vu.nl/~dijkstra/research/papers/2007connectible.pdf $\endgroup$
    – BS.
    Aug 19, 2010 at 22:04
  • $\begingroup$ More precisely, it embeds $F$ as a (connected) $G_\delta$ set $X$ in the Hilbert cube $Q=[0,1]^\mathbb{N}$. Then the closure $Y=\overline{X}$ has the property that $Y\setminus X=\bigcup_n F_n$, an increasing union of closed subsets of $Y$ (hence of $Q$). Then, if $d$ is a compatible metric on $Q$, a complete compatible metric on $X\simeq F$ is obtained by embedding it as a closed subset of $P=Q\times]0,\infty[^\mathbb{N}$, via $x\mapsto (x,(1/d(x,F_n))_n)$, and defining a complete compatible metric on $P$ (which is easy). PS: I am not Bill Johnson. $\endgroup$
    – BS.
    Aug 20, 2010 at 9:36

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