Proving a set is bounded. Let  $E = \{x \in\mathbb R^p : \sum_{i = p}^p X_i^2/\alpha_i^2 \leq 1\}$
Prove that $E$ is closed and bounded. 
To prove that $E$ is closed I used the fact that the boundary of the set $E$ is equal to $\{x \in\mathbb R^p : \sum_{i = p}^p X_i^2/\alpha_i^2 = 1\}$ and the boundary is contained in the $E.$ So $E$ is closed.
However I do not know how to prove the fact that it is bounded.
 A: To see that the set
$E = \left \{ x = (X_1, X_2, \ldots X_p) \in \Bbb R^p \mid \displaystyle \sum_1^p \dfrac{X_i^2}{\alpha_i^2} \le 1 \right \} \tag 1$
is bounded, let
$\mu = \max \{ \vert \alpha_i \vert, \; 1 \le i \le p \}; \tag 2$
then
$\dfrac{1}{\mu^2} \le \dfrac{1}{\vert \alpha_i \vert ^2} = \dfrac{1}{\alpha_i^2}, \; 1 \le i \le p, \tag 3$
whence
$\dfrac{X_i^2}{\mu^2} \le \dfrac{X_i^2}{ \vert \alpha_i \vert^2} = \dfrac{X_i^2}{\alpha_i^2}, \; 1 \le i \le p; \tag 4$
then for $x = (X_1, X_2, \ldots, X_p) \in E$ we have
$\displaystyle \sum_1^p \dfrac{X_i^2}{\mu^2} \le \sum_1^p \dfrac{X_i^2}{\alpha^2} \le 1, \tag 5$
and thus, upon multiplication by $\mu^2$,
$\displaystyle \sum_1^p X_i^2 \le \mu^2, \tag 6$
which shows that every $x \in E$ lies in the sphere of radius $\mu$ centered at the origin; thus $E$ is a bounded set.
As for $E$ being closed, the easiest approach here, I think, is to show that $\bar E$, the complement of $E$, is open; to see this, we use the fact that the function
$f(x) =  \displaystyle \sum_1^p \dfrac{X_i^2}{\alpha_i^2} \tag 7$
is continuous, which I think is quite evident; then clearly
$E = \{x \in \Bbb R^p \mid f(x) \le 1 \} = f^{-1}([0, 1]), \tag 8$
so
$\bar E = \{x \in \Bbb R^p \mid f(x) >  1 \} = f^{-1}((1, \infty)); \tag 9$
since the inverse image of an open set under a continuous mapping is open, we see that $\bar E$ is open in $\Bbb R^p$, whence $E$ is closed, being the complement of the open $\bar E$.  
A: Let $\alpha = max(\alpha_i)$ then if E is as you previously described it, then let $E_2$ be the ball of radus equal to the length of the major semi-axis ( defined as $max(\alpha_i)$. Clearly E is a subest of $E_2$ so it is then bounded
A: Try to get a lower bound on $\sum\frac{x_i^2}{\alpha_i^2}$ involving the Euclidean norm of $x$. For example:
$$\sum\frac{x_i^2}{\alpha_i^2}\geq\min_i\left(\frac1{\alpha_i^2}\right)||x||_2^2$$
