Example 5, Sec. 24 in Munkres' TOPOLOGY, 2nd ed: Is this map always continuous? Let $(X, \Vert \cdot \Vert)$ be a given normed space that has elements other than the zero vector $\theta_X$. And let $T \colon X-\{\theta_X \} \to X$ be defined by 
$$T(x) \colon= \frac{1}{\Vert x \Vert} x \ \ \ \mbox{ for all } \ x \in X, \ x \neq \theta_X.$$
Then how to determine if $T$ is continuous? Is $T$ always continuous or discontinuous? 
If $X = \mathbb{R}^n$, then $T$ is of course continuous. Right? 
After some thought: 
Let $x$ be an arbitrary element of $X-\{\theta_X\}$. Let $x_n$ be a sequence in $X-\{\theta_X\}$ such that $x_n$ converges to $x$. Then the sequence $\Vert x_n \Vert$ of non-zero real numbers converges in the usual metric space $\mathbb{R}$ to the non-zero real number $\Vert x \Vert$. So the sequence $\frac{1}{\Vert x_n \Vert}$ of reciprocals converges in $\mathbb{R}$ to $\frac{1}{\Vert x \Vert}$. Therefor, the image sequence $T(x_n)$ converges in $X$ to $T(x)$. Hence $T$ is continuous at $x$. 
Is there anything wrong about this reasoning? 
 A: $T$ is obtained from continuous functions by composition and pairing and hence is continuous. To be more precise, $T$ is:
$$X_{\neq0} \stackrel{\Delta}{\longrightarrow} X_{\neq0} \times X_{\neq0} \stackrel{\langle\|\cdot\|, \mbox{id}\rangle}\longrightarrow \mathbb{R}_{\neq0} \times X_{\neq0} \stackrel{\langle(\cdot)^{-1}, \mbox{id}\rangle }{\longrightarrow} \mathbb{R} \times X_{\neq0} \stackrel{\times}{\longrightarrow} X$$
where $\Delta$ is the diagonal map ($x \mapsto \langle x, x \rangle$) and where $\times$ over the final arrow is scalar multiplication (which is continuous by virtue of the normed space axioms).
In more detail: if $A$, $B$. $C$, and $D$ are topological spaces and $f: A \rightarrow B$,  $g:C \rightarrow D$ and $h : B \rightarrow C$ are continuous functions, then the function $\langle f, g\rangle:A \times C \rightarrow B \times D$ that maps $(a, c)$ to $(f(a), g(c))$ is continuous (pairing) as is the function $h \circ f: A \rightarrow C$ that maps $a$ to $h(f(a))$ (composition). For any topological space, $A$ the diagonal map $\Delta:A \rightarrow A \times A$ that maps $a$ to $(a, a)$ is continuous.  For a normed space $X$ over the real field, the scalar multiplication mapping $\times:\mathbb{R} \times X \rightarrow X$ that maps $(\lambda, v)$ to $\lambda v$ is continuous (the proof is an exercise in the use of the triangle inequality). Putting this together, the composition:
$$
x \mapsto (x, x) \mapsto (\|x\|, x) \mapsto (\|x\|^{-1}, x) \mapsto \|x\|^{-1}x
$$
is continuous, provided we require $x\neq0$, so that we avoid the discontinuity of $\lambda \mapsto \lambda^{-1}$ when $\lambda=0$.
A: The map $T$ takes any $x \in X \setminus \{0\}$ to the unit sphere in $X$. Furthermore, it maps the entire ray $\{ r x \mid r>0 \}$ to the same point $T(x)$. It is easy to see that the preimage of any open set in $X$ will either be the empty set or it will be a union of open wedges, so $T$ is continuous.
