Topology of uniform convergence

Remarks

The cut-off at $1$ makes $\bar d$ finite even for unbounded maps; it does not change which sequences converge. A sequence $(f_n)$ converges to $f$ in this topology if and only if $f_n \to f$ uniformly on $X$. On bounded maps, the sup metric $\sup_x d(f(x), g(x))$ gives the same topology. If $(Y, d)$ is complete, then $(Y^X, \bar d)$ is complete. If $X$ is a topological space, the set $C(X, Y)$ of continuous maps is closed in $Y^X$, since a uniform limit of continuous maps is continuous.

Example: on $\R^{[0, 1]}$, let $f_n(x) = x / n$. Then $\bar d(f_n, 0) \le 1/n$, so $f_n \to 0$ in this topology.

Non-example: let $f_n(x) = x^n$ on $X = [0, 1)$ and $Y = \R$. Then $f_n \to 0$ pointwise, but $\bar d(f_n, 0) = 1$ for every $n$, since $x^n$ takes values close to $1$ near $x = 1$. So $(f_n)$ converges in the topology of pointwise convergence, but not in the topology of uniform convergence.