Let $X$ be a topological space. Then we can define $$C_0(X):=\{f \in C(X)\mid \forall \epsilon > 0: \exists K \subseteq X \mathrm{\ compact}: \forall x \notin K: |f(x)| < \epsilon\}$$
If $X$ is locally compact, I have also seen the following definition, if $X$ is locally compact: $$C_0'(X) = \{f \in C(X)\mid \forall \epsilon > 0: \{x \in X: |f(x) | \geq \epsilon\} \mathrm{\ compact}\}$$
What is the relation between these two definitions of $C_0?$ Clearly $C_0'(X) \subseteq C_0(X)$. Do we have equality? Why do we require that $X$ is locally compact in the second definition?
Yes, the two definitions are equivalent, even without assuming local compactness. (A usual terminology to describe this behavior is to say that such a function vanishes at infinity.)
To see that $C_0'(X)\subseteq C_0(X)$, take $f\in C_0'(X)$. Then, $K_{\varepsilon}\equiv\{x\in X\,|\,|f(x)|\geq\varepsilon\}$ is compact for every $\varepsilon>0$. Use this $K_{\varepsilon}$ to verify that $f\in C_0(X)$.
Conversely, suppose that $f\in C_0(X)$ and fix an arbitrary $\varepsilon>0$. There must exist some compact $K\subseteq X$ such that $x\in X\setminus K$ implies $|f(x)|<\varepsilon$. Therefore, the set $E\equiv\{x\in X\,|\,|f(x)|\geq\varepsilon\}$ is a subset of $K$. But $E$ is closed due to the continuity of $f$ and a closed subset of a compact set is compact in any topological space. The conclusion is that $E$ is compact, which implies that $f\in C_0'(X)$.