Little-O Implies Big-O

Theorem
Let $X$ be a topological space.

Let $V$ be a normed vector space over $\R$ or $\C$ with norm $\left\Vert{\,\cdot\,}\right\Vert$

Let $f,g:X\to V$ be functions.

Let $x_0\in X$.

Suppose $f=o(g)$ as $x\to x_0$, where $o$ denotes little-O notation.

Then $f=O(g)$ as $x\to x_0$, where $O$ denotes big-O notation.

Proof
From the definition of little-O notation, we have:
 * There exists a neighborhood $U$ of $x_0$ such that $\Vert f(x)\Vert \leq \Vert g(x)\Vert$ for all $x\in U$.

By definition of big-O notation, $f=O(g)$ as $x\to x_0$.