Here is the problem link (where the title is exactly what we want to prove)
My question is:
What topology are we using in solving this problem? could anyone explain this for me please?
Here is the problem link (where the title is exactly what we want to prove)
My question is:
What topology are we using in solving this problem? could anyone explain this for me please?
The topology used is the one induced by the $1$-norm i.e. $\Vert f \Vert_1 := \int_{[0,1]} \vert f(x) \vert dx$.
This is well defined since $[0,1]$ has finite Lebesgue measure and thus $L^2([0,1]) \subseteq L^1([0,1])$.
That means, that the topology is generated by the family $\{B_{r}(f): r > 0, f \in L^1([0,1])\}$, where
$$B_{r}(f) := \{g \in L^1([0,1]) : \Vert f-g\Vert_1 < r\} ~~.$$
Hence a set $U \subseteq L^1([0,1])$ is open if and only if $U = \cup_{i \in I} B_{r_i}(f_i)$ where $I$ is some index set.