Question: 1. $0\in\{\emptyset\}$
2. $\emptyset\in\{0\}$
3. $\{0\}\subset\emptyset$
4. $\{0\}\in\{0\}$
5. $\{0\}\subset\{0\}$

I think the answers of the above questions are F, F, F, T, F. Is it wrong?

Answer: 
It does *not* follow that for groups which are not finitely generated the notion of hyperbolicity makes sense.

What follows, instead, is that for any pair $(G,S)$ such that $G$ is a group and $S$ is a generating set, the notion of hyperbolicity is well-defined: the Cayley graph $\Gamma(G,S)$ is certainly defined (and it is connected), and hyperbolicity of $\Gamma(G,S)$ certainly makes sense.

The trouble is that without any control on the cardinality of $S$, it is possible to have a group $G$ and two generating sets $S\_1,S\_2$ such that $\Gamma(G,S\_1)$ is hyperbolic whereas $\Gamma(G,S\_2)$ is not.

In fact, here's an example: $G = \mathbb Z^2$ with $S\_1 = \{(\pm 1,0),(0,\pm 1)\}$ and with $S\_2 = \mathbb Z^2$. The Cayley graph $\Gamma(G,S\_1)$ is not hyperbolic whereas $\Gamma(G,S\_2)$ is bounded and is therefore hyperbolic. That's an example where $G$ is in fact finitely generated, but one can easily cook up examples where $G$ is not finitely generated, e.g. $G = \mathbb Z^\infty$ with the "standard" generating set $S\_1 = \{\pm e\_i \mid i \ge 1\}$ is not hyperbolic, but with the generating set $S\_2=G$ it is hyperbolic.

What happens in the finitely generated case is that if $S\_1,S\_2$ are two **finite** generating sets for the same group $G$, then $\Gamma(G,S\_1)$ and $\Gamma(G,S\_2)$ are quasi-isometric to each other, and hyperbolicity is a quasi-isometry invariant, so $\Gamma(G,S\_1)$ is hyperbolic if and only if $\Gamma(G,S\_2)$ is hypebolic. Thus one says that a finitely generated group $G$ is hyperbolic if and only if some (hence any) finite generating set $S$ results in a hyperbolic Cayley graph $(G,S)$.

The summary statement is that hyperbolicity of a **finitely generated** group is well-defined independent of the choice of **finite** generating set, but take away those bold faced words and the statement becomes false.

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**ADDED:** In answer to a followup question in the comments, every Gromov hyperbolic space (i.e. every geodesic metric space satisfying the thin triangle property) has a Gromov boundary.

And, given a group $G$ and two generating sets $S\_1$, $S\_2$, if $\Gamma(G,S\_1)$ and $\Gamma(G,S\_2)$ are both hyperbolic, then both of them do have **a** Gromov boundary. And if $S\_1,S\_2$ are both finite then one can use the quasi-isometry between $\Gamma(G,S\_1)$ and $\Gamma(G,S\_2)$ to deduce that their Gromov boundaries are homeomorphic; so in the finitely generated case you are free to speak about **THE** (homeomorphism class of the) Gromov boundary.

However, without the assumption that $S\_1,S\_2$ are both finite, it need not be true that the Gromov of $\Gamma(G,S\_1)$ is homeomorphic to the Gromov boundary of $\Gamma(G,S\_2)$.

For a counterexample, take your favorite infinite, finitely generated, hyperbolic group $G$. With any finite generating set $S\_1$, the Gromov boundary is nonempty. With the infinite generating set $S\_2=G$, the Gromov boundary is empty.

Anyway, it follows that to speak of **THE** Gromov boundary of $G$ is nonsensical in the infinitely generated case.
If you read Gromov's original book, "hyperbolic group" means a group endowed with a left-invariant metric, that is hyperbolic. Quite equivalently, it's a group endowed with an isometric action on a hyperbolic space. The boundary can be defined in this broad context. For instance, if $\mathbf{Z}$ acts on the hyperbolic plane by a parabolic isometry, its boundary is a singleton.

He then especially studied the case of finitely generated groups with a word metric (in which case the boundary is independent of such a choice).

A more general realm is that of locally compact groups: if the word metric with respect to some compact generating subset is hyperbolic, then the boundary doesn't depend on the particular choice. This includes a rich supply of examples that do not resemble the finitely generated ones, e.g., connected Lie group with left-invariant negatively curved metric and millefeuille spaces (below the boundary of a millefeuille space, borrowed from [here](https://arxiv.org/abs/1212.2229); it purports to represent the one-point compactification of $\mathbf{Q}\_p\times\mathbf{R}$).

[![](https://i.stack.imgur.com/BDK0v.jpg)](https://i.stack.imgur.com/BDK0v.jpg)

There's even broader settings where there's a canonical such metric, in the context of Polish groups (see Rosendal's work), but I'm not aware it's ever been studied in the angle of hyperbolicity.