Define the metric $d(f,g)\triangleq \sup\_{x \in [0,1]} \|f(x)-g(x)\|$ on the set $\operatorname{B}$ of uniformly bounded functions from the interval $[0,1]$ to $\mathbb{R}$, fix $g \in \operatorname{B}$, and define the map $F:\operatorname{B}\rightarrow [0,\infty)$ by $F(f):=d(g,f)$. Is the map $F$ continuous? It certainly is on the subset $C([0,1],\mathbb{R})$ but what about on the rest of the space?



Uniform distance from a discontinuous function is continuous
If $d$ is a metric on $B$ then the mapping $F(f) := d(g,f)$ is certainly continuous with respect to the topology induced by the metric $d$:


Let $(f\_n)\_{n\in\mathbb{N}}$ a sequence in $B$ that converges to $f \in B$ w.r.t. $d$. This is equivalent to
$$
d(f,f\_n) \to 0.
$$
Hence, by the triangular inequality
$$
F(f\_n) = d(g,f\_n) \leq d(g,f) + d(f,f\_n) \to d(g,f) + 0 = F(f).
$$
On the other hand by the reversed triangular inequality
$$
F(f\_n) = d(g,f\_n) \geq d(g,f) - d(f,f\_n) \to d(g,f) + 0 = F(f).
$$
This means
$$
F(f) \leq \lim\_{n\in\mathbb{N}} F(f\_n) \leq F(f)
$$
which implies the continuity of $F$.


