Finding the topological boundary of an embedding 
Let $M = \mathbb{R}^n \setminus \{0\}$ and $\phi : M \to \mathbb{R}^n$ defined by $\phi(x_1, .., x_n) = \frac{r+1}{r}(x_1, .., x_n)$ be an embedding, where $r = \sqrt{\sum_{i=1}^n x_i^2}$.

Now the paper I'm reading claims that the boundary of $\phi[M] \subseteq \mathbb{R}^n$ is $\mathbb{S}^{n-1}$ centered at the origin, but I don't see how that's possible. If we had $\phi(x) = \frac{x}{|x|}$ where $x \in M$, then I'd definitely agree that $\text{Bd}(\phi(M)) = \mathbb{S}^{n-1}$, but in it's current form we have $\phi(x) = x + \frac{x}{|x|}$ for $x \in M$.
Is there an error in the paper?

$\phi (x)=(1+\frac {1}{\|x\|})x.$ 
(i). Verify that the image $\phi (M)=\{y:\|y\|>1\} .$ 
(ii). Since $\phi (M)$ is open its boundary  is $\overline {\phi (M)}\setminus \phi (M),$ which is a subset of $\{v:\|v\|\leq 1\}.$ 
For $\|v\|<1$ check that $v\ne \overline {\phi (M)}.$ 
For $\|v\|=1$ we have $\lim_{m\to \infty} \|\phi (mv)-v\|=0$ so $v\in \overline {\phi (M)}$. So $v$ is in the boundary. So the boundary is $\{v:\|v\|=1\}=S^{n-1}.$
