# How thinly connected can a closed subset of Hilbert space be?

Let H be a separable (and infinite-dimensional) Hilbert space. Is it known whether there exists an infinite subset C of H with the following properties.? (1) C is connected and closed in H. (2) No infinite proper subset of C is both connected and closed in H.

Perhaps the answer depends upon whether or not the Axiom of Choice is assumed to hold. In any case, no finite-dimensional Euclidean space-because it is locally compact-can contain a subset with these properties.

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Hilbert space certainly has lines in it, if that's what you mean. – some guy on the street May 18 '10 at 20:15
..and lines certainly have closed intervals in them ;) Garabed: what's the analog result for H= R^n ? – Pietro Majer May 18 '10 at 20:26
IIRC the universal properties of $\ell_2$ (in the realm of Polish spaces) guarantee that if there is a Polish space with these properties then there is one which is a closed subspace of $\ell_2$. – François G. Dorais May 18 '10 at 21:57
I did remember correctly, see Theorem 4.17 and the following remark in Kechris's Classical Descriptive Set Theory. – François G. Dorais May 19 '10 at 1:36
To Pietro Majer: One analog is-If E(M,p,e) denotes the set of all points of M lying within any positive distance e of any point p belonging to an infinite closed connected subset M of R^n, then E(M,p,e) contains an infinite connected subset. This theorem does not hold in an infinite dimensional and separable Hilbert space. – Garabed Gulbenkian May 31 '10 at 19:49

This is not really an answer, just remarks on the problem, but it would be too long for a comment. Apologies.

As remarked by François Dorais in a comment, you can phrase the question entirely in terms of polish spaces (i.e. completely metrizable and separable topological spaces). And since a connected metric space with at least two points is infinite, you are asking if there exists an infinite connected polish space such that any proper closed subspace is totally disconnected (equivalently, any proper closed connected subset is a point, or empty).

A first (failed) attempt to an example would be the complete Erdös space $E_c$ (Erdös, Annals of Math vol 41 1940), defined as the subspace of $\ell^2(\mathbb{N})$ where all coordinates are irrationals. It is polish and totally disconnected, but admits a connectification namely a (still polish) topology on $E_c\cup\{\infty\}$ that makes it connected (and of course induces the one on $E_c$). So it is rather "thinly" connected, but maybe not in your sense.

Another amazing property of $E_c$ is that it is homeomorphic to the subspace of $\ell^2(\mathbb{N})$ where all coordinates belong to $\{0\} \cup \{1/n\}_{n\geq 1}$. This sounds rather improbable.

All this (and much more) is explained in papers on JJ Dijkstra publications page e.g. 27, 30 and 32.

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2) As hinted by Garabed, such a space $X$ cannot be locally compact. Indeed, in locally compact spaces, components coincide with quasi-components. Let $F$ be a closed ball of some radius about some $x\in X$, so that $F\ne X$, and let $C$ be the component of $x$ in $F$. If $C\ne\{x\}$, then $C$ is a closed connected nondegenerate subset of $X$. If $C=\{x\}$, then $\{x\}$ is also a quasi-component, so $x$ is contained in arbitrarily small clopen sets in $F$. Then they are also clopen in $X$, so $X$ cannot be connected.
3) There exists a connected Polish space $X$ containing no nondegenerate compact connected subspace. For instance, the graph of $f(x)=\sum_{n=1}^\infty 2^{-n}\sin(\frac1x-r_n)$, where $r_n$ is the $n$th rational number (in some order) and $\sin(\infty)=0$. See Kuratowski's "Topology" (volume II, section 47.IX in the 1968 edition). In fact, such an $X$ can even be locally connected (of course, it cannot be locally path-connected at any point).
4) The graph of a discontinuous function $f:\Bbb R\to\Bbb R$ satisfying $f(x+y)=f(x)+f(y)$ can be connected, and in that case it contains no nondegenerate bounded connected subset.