Suppose that $X$ is a topological space and $Y$ a metric space.

In the special case of functions on the complex plane where it is often used, the definition can be given as follows.

As an example we can look at the set ${\mathcal{F}}$ of entire functions where $f(z)=z^{2}+t$ for any $t\in[0,1]$. Obviously each such $f$ is unbounded itself, however if we take a small neighbourhood around any point we can bound all $f\in{\mathcal{F}}$. Say on an open ball $B(z_{0},1)$ we can show by triangle inequality that $\lvert f(z)\rvert\leq(\lvert z_{0}\rvert+1)^{2}+1$ for all $z\in B(z_{0},1)$. So this set of functions is locally bounded.

Another example would be say the set of all analytic functions from some region $G$ to the unit disc. All those functions are bounded by 1, and so we have a uniform bound even over all of $G$.

As a counterexample suppose the we take the constant functions $f_{n}(z)=n$ for all natural numbers $n$. While each of these functions is itself bounded, we can never find a uniform bound for all such functions.