Vector-Valued Functions and Continuity Why is it that when a vector-valued function $r(t)$ is continuous at some time $t$ then $\|r(t)\|$ is also continuous at that time $t$, but the converse is not true? That if $\|r(t)\|$ is continuous at time $t$, then $r(t)$ is discontinuous at time $t$.
 A: It is due to the fact that the norm is a continuous function and the composition of continuous functions is continue. $t\rightarrow\|r(t)\|$ is the compostion of $r$ and the norm.
On the other side, if the $f\circ g$ is continuous with $f$ continuous, you can't conclude that $g$ is continue. For example the composition $c\circ f$ is always continuous where $c$ is a constant function.
A: First of all, you should be careful with your phrasing. It's false that $\| r\|$ is continuous at $t$ implies $r$ is discontinuous at $t$. I believe what you mean is that $\|r\|$ continuous at $t$ does not imply $r$ continuous at $t$, which is true.
In fact, it's even true for the reals. Consider the function $f: \mathbb{R} \to \mathbb{R}$ where
$$f(x) = \begin{cases}  1&\text{ x is rational}\\
-1 & \text{else}\end{cases}
$$
Then $|f(x)|$ is continuous everywhere, but $f$ is continuous nowhere.  
To extend this to the vector valued case, just take a vector valued function that looks like $f$ in the first coordinate, and $0$ elsewhere. 
