Why is this curve not closed? Consider the curve $$\gamma (t) =\left( \cos(t^3+t), \sin(t^3+t) \right) $$
I am asked to show that a reparamaterization of a closed curve is not necessarily  closed. The book provides this as a counter example. Why is this curve not closed? 
I think I am having some trouble showing curves are open/closed. 
 A: Let $\gamma(t) = (cos (t^3+t), sin (t^3+t))$ and assume that there were such a $T \ne 0$ that made $\gamma(t+T) = \gamma(t)$. Then it would be true for its derivative too (ie $\gamma'(t+T) = \gamma'(t)$).
This would especially mean that $\gamma'(T) = \gamma'(0)$, but $\gamma'(t) = (3t^2+1)(-sin(t^3+t), cos(t^3+t))$. But $|\gamma'(0)| = 1$, but $|\gamma'(T)| = 3T^2+1$, and since $T \ne 0$ then $\gamma'(0) \ne \gamma'(T)$.  
A: Here are two attempts to guess what your book is after.
If the book really gives no more context and no interval of definition, I find this example confusing.
A formula alone is not a curve; a curve is a function that can be defined using a formula, but also a domain is needed (and it is not implicitly understood clearly enough in this case).
Attempt 1.
The curve $\sigma:[0,2\pi]\to\mathbb R^2$ given by $\sigma(t)=(\cos(t),\sin(t))$ is closed.
The mapping $\phi:\mathbb R\to\mathbb R$, $\phi(t)=t+t^3$ is a valid change of parametrization, so $\sigma\circ\phi$ is a reparametrization of $\sigma$.
In fact, $\sigma\circ\phi=\gamma$ (the formula you gave), but $\gamma:[0,2\pi]\to\mathbb R^2$ is not a closed curve.
This is a bit silly, though, since the domain of $\gamma$ should be $\phi^{-1}([0,2\pi])$, not $[0,2\pi]$.
A reparametrization should also change the interval of definition.
Attempt 2.
It might be that your book requires that a closed curve $\gamma:[a,b]\to\mathbb R^n$ satisfies not only $\gamma(a)=\gamma(b)$ but also $\gamma'(a)=\gamma'(b)$.
This only makes sense if you require $\gamma$ to be differentiable in the first place; you did not mention about differentiability assumptions.
If $T$ is the number satisfying $T+T^3=2\pi$, then your $\gamma$ defined on $[0,T]$ does not satisfy the second condition.
I find this silly, too, since if matching derivatives are required at endpoints, then the definition of reparametrization should include conditions on the derivatives at the endpoints to remove this issue.
