Quasi-isometries vs Cayley Graphs

The following questions might be trivial, however, I couldn't solve them:

Let $G$ be generated by a finite symmetric set $S.$ Suppose that $\Gamma(G,S)$ is the corresponding right Cayley graph of $G.$ $X$ is a metric space(or, maybe a topological space with some nice structure).

(1) Is there a way to check the following property of a space $X:$ $X$ is not quasi-isometric to a space $Z$ which is quasi-isometric to a(hence, every) Carley graph $\Gamma(G,S)$ of some f.g group $G.$

I.e, If we partition the space of spaces upto quasi-isometric equivalence then does every equivalence class contain a space which is quasi-isometric to a Cayley graph of some f.g group $G?$

(2) By Stalling's theorem, # of ends is a geometric property of the group. Does this mean that # of ends is a quasi-isometric invariant of the spaces which are quasi-isometric to Cayley graphs?

If the answer of question #2 is affirmative and equivalence class question above fails; i.e, there is an equivalence class whose elements are not quasi-iso. to any Cayley graph ; then what is the example of spaces $W_1, W_2$ which are not quasi-iso. to any Cayley graph but $W_1$ is quasi-isometric to $W_2$ ,however, # of ends of $W_1$ is different than the # of ends of $W_2.$

Thank you.

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There are a lot of conditions missing in order for these questions to make a little sense. Make the spaces geodesic, "almost homogeneous" and of bounded geometry for a start. – Theo Buehler Feb 25 2011 at 7:43
Could you precise what definition of the number of ends of a metric space you use? – Benoît Kloeckner Feb 25 2011 at 7:56
@Buehler: Yes, I couldn't figure out the minimal conditions on $X.$ I am interested in the invariance of number of ends. And the end is the number of components of $X-B(n)$ as n->\infity. I know that it doesn't make sense for general metric spaces. But, according to answer given below, this number should make sense for a large collection of spaces also. – Niyazi Feb 25 2011 at 9:32
@Niyazi: the problem is that the number of ends defined in your way is infinite for an unbounded discrete space (in particular it sis far from being a quasi-isometric invariant, even for spaces quasi-isometric to a Cayley graph), so you should really make explicit the conditions on $X$, as asked by Theo Buehler. – Benoît Kloeckner Feb 25 2011 at 15:43
Niyazi, Stallings's Theorem does not say that the number of Ends is a quasi-isometric invariant of the group. It gives a condition under which a fg group has infinitely many ends. – HW Feb 25 2011 at 16:07

I guess that a star (a tree with $n$ infinite branches issued from a single vertex) should answer at least your first question. It should have $n$ ends, whatever meaningful definition you use, an we know that a group has $1$, $2$ or an infinity of ends.

Since quasi-isometry is an equivalence relation, you do not need to invoke a space $Z$ in your first question and the answer of your second question is obviously positive.

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My comment was intended to exclude such examples as you give in the first paragraph (among other things). With any of the definitions I know, $n$ parallel lines in $\mathbb{R}^2$ with the Euclidean metric will have $2n$ ends. – Theo Buehler Feb 25 2011 at 8:24
@Kloeckner: So, if $Z$ is quasi-isometric to 5-star then can we conclude that Z has 5 ends? I know that the number of ends is a quasi-iso. invariant only for cayley graphs, is it again an invariant of non-Cayley but quasi-isometric spaces? Also, according to your answer examples of $W_1$ and $W_2$ exist. Is it correct? Thank you. – Niyazi Feb 25 2011 at 9:24
Benoit answers both questions. E.g. the union of the $x$ and $y$ axes in the plane is a metric space with 4 ends. It is therefore not q-i to any group (which must have 0, 1, 2, or infinity ends). It is also not q-i to the union of the $x$, $y$, and $z$ axes in $R^3$, which has 6 ends. In other words, yes, the number of ends is a q-i invariant of metric spaces, not just of groups. (It is easy to prove: a q-i induces a bijection on the ends.) – aaron Feb 25 2011 at 12:45
@aaron: Either you're working with a coarse notion of ends which I don't know or you're a bit simplistic here. I agree with your argument for proper geodesic spaces. Could you please elucidate your argument with a precise definition of your notion of an end? With the usual topological definition of an end, a quasi-geodesic doesn't need to lie in a single end, see my first comment to this answer. – Theo Buehler Feb 25 2011 at 14:10
Oh right, certainly one needs a coarse notion of ends. The integers should have 2 ends regardless of whether or not you draw in the edges connecting $n$ to $n+1$. – aaron Feb 25 2011 at 21:39