Can it be shown that $\|x\|_2 \to \infty \implies \|x - y\|_2 \to \infty$?? Let $y$ be a constant vector. 
By triangle inequality, $\|x - y\|_2 \to \infty$ implies that $\|x\|_2 \to \infty$. 
Can we show the reverse, $\|x\| \to \infty$ implies $\|x - y\|_2 \to \infty$?
I've tried using reverse triangle inequality, but it is not working out. Please help!
 A: Just $$\|x-c\|+\|c\|\geq\|x\|$$
A: For any norm $\Vert \cdot \Vert$ and any vectors $u$, $v$, the triangle inequality states that
$\Vert u + v \Vert \le \Vert u \Vert + \Vert v \Vert; \tag{1}$
thus
$\Vert u \Vert = \Vert (u - v) + v \Vert \le \Vert u - v \Vert + \Vert v \Vert; \tag{2}$
reversing the roles or $u$ and $v$ yields
$\Vert v \Vert = \Vert (v - u) + u \Vert \le \Vert v - u \Vert + \Vert u \Vert; \tag{3}$
(2) and (3) are respectively equivalent to
$\Vert u \Vert - \Vert v \Vert \le \Vert u - v \Vert\ \tag{4}$
and
$\Vert v \Vert - \Vert u \Vert \le \Vert v - u \Vert\; \tag{5}$
since $\Vert u - v \Vert = \Vert v - u \Vert$, (4) and (5) together imply
$\vert \Vert u \Vert - \Vert v \Vert \vert \le \Vert u - v \Vert; \tag{6}$
now taking $u = x$ and $v = c$ we find
$\vert \Vert x \Vert - \Vert c \Vert \vert \le \Vert x - c \Vert; \tag{7}$
as $\Vert x \Vert \to \infty$, (7) implies $\Vert x - c \Vert \to \infty$ as well.
One can restrict oneself to $\Vert \cdot \Vert_2$ if so desired, but the result holds for any norm.
