If $G$ is a finitely-generated group, then we say that $G$ is Gromov-Hyperbolic if it's Cayley Graph, $\operatorname{Cay}(G, S)$, is a Gromov-Hyperbolic metric space.
Now in the case of the group of integers under addition $\mathbb{Z} = \langle g \rangle$, we have that $\operatorname{Cay}(\mathbb{Z}, \{g\})$ is isometric to $\mathbb{R}$ and we know that $\mathbb{R}$ is not a Gromov-Hyperbolic metric space, so it follows that $\mathbb{Z}$ is not Gromov Hyperbolic.
However, I have seen in some notes on Geometric Group Theory that $\mathbb{Z}$ is Gromov Hyperbolic. Why is this so? And what is the error I'm making?
As metric spaces both $\mathbb{Z}$ and $\mathbb{R}$ are Gromov $0$-hyperbolic as can be easily seen from the Gromov four-point condition:
$$(x|z)_o \geq \min((x|y)_o,((y|z)_o) - \delta,$$
with $(x|y)_o = \frac{1}{2}(d(x,o) + d(y,o) - d(x,y))$. This has to hold for any $x,y,z,o \in X$ for your space $X$ to be $\delta$ hyperbolic. Here $\delta = 0$.
Note that in $\mathbb{R}$ and $\mathbb{Z}$ the triangles are degenerate and we have the metric $d(x,y) = |x-y|$. Plugging this information in to the above four-point condition, we can see that it is satisfied for all points with $\delta = 0$.