Show that $\|x(t) - x_o\| \to \infty $ if $ \|x(t) \| \to \infty$

Question

Let $\| \cdot \|$ be any norm on $\mathbb{R}^n$, $x_o$ a vector on $\mathbb{R}^n$

Show that $\|x - x_o\| \to \infty$ if $\|x \| \to \infty$

This seems intuitively obvious, but is there a good, rigorous way of showing this?

I am thinking of some property that shows "$\|x(t) - x\| \geq \|x(t)\|\|x_o\|$" so taking $\|x(t)\|$ to infinity yields the conclusion as required.

Can't seem to think of such property. Can anyone help?

Answer 1

By the triangle inequality we have $$ ||x||=||x-x_0+x_0||\leq ||x-x_0||+||x_0|| $$ for all $x$, and this can be rearranged to $$ ||x-x_0||\geq ||x||-||x_0|| $$ Therefore if $x_0$ is fixed and $||x||\to\infty$, then $||x-x_0||\to\infty$.

Answer 2

use $||x-x_0||\geq||x||-||x_0||$