# Uniform distance from a discontinuous function is continuous

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?

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 $$F$$.